How much energy must be supplied to break a single 21Ne nucleus into separated protons and neutrons if the nucleus has a mass of 20.98846 amu?

What is the nuclear binding energy for 1 {\rm mol} of {\rm ^{21}Ne}?

Answers

Answer 1

Answer:

1)There is 2.68 * 10^-11 J of energy needed

2) The nuclear binding energy for 1 mol of Ne is 1.6 *10^13 J/mol

Explanation:

Step 1: Data given

The nucleus of a21Ne atom has a amass of 20.98846 amu.

Step 2: Calculate number of protons and neutrons

The number of electrons and protons in an 21Ne atom = 10

The number of neutrons = 21 -10 =11

Step 3: mass of the atom

Mass of a proton = 1.00727647 u

Mass of a neutron = 1.0086649 u

The mass of the atom = mass of all neutrons + mass of protons

Mass of atom = 11*1.0086649 + 10*1.00727647  = 21.1680786 amu

Step 4: Calculate change of mass

The change in mass = Mass of atom - mass of neon

Δmass = 21.1680786 - 20.98846

Δmass = 0.1796186

Step 5: Calculate mass for a single nucleus

The change of mass for a single nucleus is = Δmass / number of avogadro

Δmass of nucleus = 0.1796186 / 6.022*10^23

Δmass of nucleus =2.98 * 10^-25 grams = 2.98 * 10^-28 kg

Step 6: Calculate energy to break a Ne nucleus

Calculate the amount of energy to break a Ne nucleus

ΔEnucleus = Δmass of nucleus * c²

⇒ with c = 2.9979 *10^8 m/s

ΔEnucleus = 2.98 * 10^-28 kg * (2.9979*10^8)² = 2.68 * 10^-11 J

What is the nuclear binding energy for 1 mol of Ne?

ΔE= ΔEnucleus * number of avogadro

ΔE= 2.68 * 10^-11 J * 6.022*10^23

ΔE= 1.6 *10^13 J/mol


Related Questions

Linoleic acid is a polyunsaturated fatty acid found, in ester form, in many fats and oils. Its doubly allylic hydrogens are particularly susceptible to abstraction by radicals, a process that can lead to the oxidative degradation of the fat or oil. The radical formed by abstraction of one of the doubly allylic hydrogens is an allylic radical that has three resonance structures. Complete one of these resonance structures by dragging bonds and electrons to their appropriate positions.

Answers

Final answer:

Linoleic acid is a polyunsaturated fatty acid that forms an allylic radical with three resonance structures when its doubly allylic hydrogens are abstracted by radicals. These types of fats are liquid at room temperature and are nutritionally significant due to the body's inability to produce them. They carry numerous health benefits, including heart health and anti-cancer properties.

Explanation:

Linoleic acid is a polyunsaturated fatty acid found in various fats and oils. The hydrogen atoms in its doubly allylic positions are readily abstracted by radicals, leading to the oxidative degradation of the fat or oil. The resulting radical, called an allylic radical, has three resonance structures. Completing one resonance structure involves correctly positioning bonds and electrons.

Linoleic acid, like other polyunsaturated fats, can have more than one double bond in its structure. These fats are usually liquid at room temperature, like canola oil. When a fatty acid has a double bond in its structure, it is classified as unsaturated. Polyunsaturated fatty acids such as omega-3 fatty acids are nutritionally important because the human body cannot manufacture them. They must be obtained from the diet.

Fatty acids such as omega-3 fatty acids have a number of health benefits. They reduce the risk of heart attacks, reduce blood triglyceride levels, decrease blood pressure, and prevent thrombosis by inhibiting blood clotting. They also have anti-inflammatory properties and may help lower the risk of certain cancers.

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5 Fe2+ + MnO4− + 8 H+ ⇄ 5 Fe3+ + Mn2+ + 4 H2O In a titration experiment based on the equation above, 25.0 milliliters of an acidified Fe2+ solution requires 14.0 milliliters of standard 0.050-molar MnO4− solution to reach the equivalence point. The concentration of Fe2+ in the original solution is…

(A) 0.0010 M
(B) 0.0056 M
(C) 0.028 M
(D) 0.090 M
(E) 0.14 M

Answers

Answer:

0.14 M

Explanation:

Volume of Fe2+ required for titration = 25 mL

Volume of MnO4- required for titration = 14 mL

Concentration of MnO4- = 0.05 M

We have to find the concentration of Fe2+.

Let it be x.

In a redox reaction:

Number of equivalence of oxidising agent = Number of equivalence of reducing agent.

Formula for number of equivalence: n*M*V

n = n-factor (number of electrons gained or lost by 1 mole of reducing or oxidising agent)

M = molarity

V = volume

n-factor of Fe2+ is 1 as it changes to Fe3+ by loosing 1 electron

n-factor of MnO4- is 5 as it changes to Mn2+ by gaining 5 electrons.

Number of equivalence of Fe2+ = Number of equivalence of MnO4-

1*(x)*25 = 5*0.05*14

x = 0.14 M

Final answer:

The concentration of Fe2+ in the original solution is 0.14 M. This is calculated using stoichiometry and the molarity of the MnO4− solution used in the reaction.

Explanation:

The given question is about stoichiometry and molarity in a chemistry experiment involving a redox reaction of Fe2+ and MnO4−. To determine the concentration of Fe2+ in the original solution, we will use the reaction stoichiometry and the information given about the volume and concentration of the MnO4− solution. From the balanced equation, 5 moles of Fe2+ react with 1 mole of MnO4−. We know that the number of moles can be calculated as the product of volume (in litres) and molarity. Hence the moles of MnO4− is 14.0 mL (or 0.014 L) * 0.050 mol/L = 0.0007 mol. Therefore, the moles of Fe2+ = 5 * moles of MnO4− = 5 * 0.0007 mol = 0.0035 mol. The volume of Fe2+ solution is 25.0 mL or 0.025 L. Therefore, the concentration (molarity) of Fe2+ in the original solution = moles/volume = 0.0035 mol/0.025 L = 0.14 M. Therefore, the answer is (E) 0.14 M.

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A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 3.70kg of water at 30.5 C. During the reaction 97.1 kJ of heat flows out of the flask and into the bath.
Calculate the new temperature of the water bath. You can assume that the specific heat capacity of water under these conditions is 4.18J / g*K. Round to the 3 significant digit.

Answers

Answer:

The new temperature of the water bath is 36.8°C

Explanation:

A Calorimety problem to apply this formula:

Q = m . C . ΔT

Where Q is heat

m is mass

C is specific heat capacity

ΔT = T° final - T° initial

Let's make some conversion before

97.1 kJ .1000 = 97100 Joules

3.70 kg . 1000 = 3700 g

This because the specific heat capacity units

ΔT it's a difference and  mathematically, in numbers, it is the same value in   °C as in K

97100 J = 3700 g . 4.18J / g°C ( T°final - 30.5°C)

97100 J =  15466 J/°C (T°final - 30.5°C)

97100 J / 15466 °C/J = (T°final - 30.5°C)

6.28°C = T °final - 30.5°C

6.28° C + 30.5°C = T°final

36.8°C = T° final

The target diol is synthesized in one step from 1-methylcyclopentene, but your lab partner exhausted the supply of that alkene. Fortunately, you have plenty of isomers (C6H10) on hand from which to synthesize 1-methylcyclopentene and, ultimately, the diol. Provide the missing reagents and organic structures needed to most efficiently produce the target product.

Answers

Answer:

The reagents are [tex]CH_{3}CH_{3}O^{-},OsO_{4},NaHSO_{3}and H_{2}O[/tex].

Explanation:

1-Methylenecyclopentene is treated with HBr form 1-bromo-1-methylcyclopentane, which is treated with strong base ethoxide ion and forms 1-methylcyclopent-1-ene.

This alkene is treated with osmium tetraoxide in the presence of sodium bisulfite to form target product.

The chemical reaction is as follows.

Final answer:

With the help of isomers of C6H10, such as cyclohexene, it's possible to synthesize 1-methylcyclopentene by methylation and subsequent dehydrohalogenation. Once 1-methylcyclopentene is produced, the diol can be synthesized via a dihydroxylation process involving an oxidizing agent.

Explanation:

The synthesis of the target diol from 1-methylcyclopentene can be achieved even if the starting alkene is not available by using isomers of C6H10. The first step involves synthesizing 1-methylcyclopentene from an appropriate isomer. For instance, cyclohexene, an isomer of C6H10, can be first transformed into 1-methylcyclohexene through an acid-catalyzed methylation reaction. To get 1-methylcyclopentene, a controlled elimination (dehydrohalogenation) step is necessary. This will produce the alkene with the double bond at the correct location. Once you have synthesized 1-methylcyclopentene, creating the diol is straightforward by applying the reaction conditions for dihydroxylation, which add a hydroxyl group to each carbon of the double bond.

Let's illustrate:

Start with cyclohexene (C6H10).Add HCl and CH3OH as reagents under acidic conditions to produce 1-methylcyclohexene.Heat this product with a strong base (e.g., KOH) to induce dehydrohalogenation and result in 1-methylcyclopentene.Finally, add an oxidizing agent (e.g., KMnO4) under acidic conditions to create the target diol.

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which of the following pieces of laboratory glassware should be used to most accurately measure out a 25.00 ml sample of a solution?
Ii. Out of a 25 mL graduated cylinder, 150 mL beaker, and a 50 mL Erlenmeyer flask which has the least uncertainty (most accurate volume)? Explain iii. If you had to accurately measure 20 mL of a liquid which of the three pieces of glassware would you use, 25 mL graduated cylinder, 150 mL beaker or a 50 mL? Explain

Answers

Final answer:

To most accurately measure out a 25.00 mL sample of a solution, a 25 mL graduated cylinder should be used. The same graduated cylinder can also be used to accurately measure 20 mL of a liquid.

Explanation:

In order to most accurately measure out a 25.00 mL sample of a solution, a 25 mL graduated cylinder should be used. A graduated cylinder is designed with calibrated markings that allow for precise measurement of volume. It typically has a smaller uncertainty or error compared to a beaker or an Erlenmeyer flask.

If you had to accurately measure 20 mL of a liquid, you would still use the 25 mL graduated cylinder. Even though the cylinder has a larger capacity, it can still accurately measure smaller volumes. It is important to note that using a piece of glassware with a larger capacity than the required volume ensures that there is no spillage during the measurement process.

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Final answer:

The most accurate piece of laboratory glassware to measure out a 25.00 mL sample of a solution would be a 25 mL graduated cylinder. If you had to accurately measure 20 mL of a liquid, the best choice among the three options would still be the 25 mL graduated cylinder.

Explanation:

The most accurate piece of laboratory glassware to measure out a 25.00 mL sample of a solution would be a 25 mL graduated cylinder. A graduated cylinder is designed to measure liquid volume with good accuracy. The markings on the cylinder allow for precise measurements, usually to the nearest 0.1 mL. By using the markings and reading the bottom of the meniscus, you can determine the volume of the solution.

If you had to accurately measure 20 mL of a liquid, the best choice among the three options would still be the 25 mL graduated cylinder. Although the graduated cylinder has a larger capacity than the required volume, it still provides accurate measurements within its range. It allows for better precision compared to a beaker or an Erlenmeyer flask.

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Sodium metal reacts with water to produce hydrogen gas and sodium hydroxide according to the chemical equation shown below.
When 0.0 25 mol of Na is added to 100.00 g of water, the temperature of the resulting solution rises from 25.00°C to 35.75°C.

If the specific heat of the solution is 4.18 J/(g · °C), calculate ΔH for the reaction, as written.

2 Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g) ΔH= ?

Answers

Answer: The enthalpy change of the reaction is -361.6 kJ

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Moles of sodium = 0.025 moles

Molar mass of sodium = 23 g/mol

Putting values in above equation, we get:

[tex]0.025mol=\frac{\text{Mass of sodium}}{23g/mol}\\\\\text{Mass of sodium}=(0.025mol\times 23g/mol)=0.575g[/tex]

We are given:

Mass of water = 100.00 g

Mass of sodium = 0.575 g

Mass of solution = 100.00 + 0.575 = 100.575 g

To calculate the amount of heat absorbed, we use the equation:

[tex]q=m\times C\times \Delta T[/tex]

where,

q = amount of heat absorbed = ?

m = mass of solution = 100.575 g

C = specific heat capacity of solution = 4.18 J/g°C

[tex]\Delta T[/tex] = change in temperature = [tex](T_2-T_1)=(35.75-25.00)=10.75^oC[/tex]

Putting all the values in above equation, we get:

[tex]q=100.575g\times 4.18J/g^oC\times 10.75^oC=4519.34J=4.52kJ[/tex]

When heat is absorbed by the solution, this means that heat is getting released by the reaction.

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

For the given chemical reaction:

[tex]2Na(s)+2H_2O(l)\rightarrow NaOH(aq.)+H_2(g)[/tex]

When 0.025 moles of sodium is reacted, the heat released by the reaction is 4.52 kJ

So, when 2 moles of sodium will react, the heat released by the reaction will be = [tex]\frac{4.52}{0.025}\times 2=361.6kJ[/tex]

Hence, the enthalpy change of the reaction is -361.6 kJ

Final answer:

To calculate ΔH for the reaction, we can use the equation ΔH = q / n, where q is the heat transferred and n is the number of moles involved in the reaction. Given the mass of the solution, specific heat, and temperature change, we can calculate the heat transferred and use it to find ΔH.

Explanation:

The given chemical reaction is:

2 Na(s) + 2 H₂O(l) → 2 NaOH(aq) + H₂(g)

To calculate ΔH for the reaction, we can use the equation:

ΔH = q / n

Where:

ΔH is the enthalpy changeq is the heat transferredn is the number of moles involved in the reaction

Given that 0.025 mol of Na is added to 100.00 g of water and the temperature rise is 10.75°C, we can calculate the heat transferred:

q = m × c × ΔT

Where:

m is the mass of the solution in grams (100.00 g in this case)c is the specific heat of the solution (4.18 J/(g · °C) in this case)ΔT is the temperature change (35.75°C - 25.00°C = 10.75°C in this case)

After calculating q, we can use it to find ΔH:

ΔH = q / n

Substituting the values, we get:

ΔH = q / n = (m × c × ΔT) / n

So, ΔH for the reaction is the calculated value.

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Which has not been suggested as a reasonably practical way to store large amounts of hydrogen in relatively small spaces for its use as a fuel?


A. Liquefy hydrogen under pressure and store it much as we do with liquefied natural gas today.


B. Absorb hydrogen onto activated charcoal; then heat the mixture to release the hydrogen.


C. Store it in the form of ionic metal hydrides, such as LiH, which release hydrogen gas when they react with water.


D. Encapsulate hydrogen molecules in fullerene molecules (large, carbon-based molecules that can act like cages) that may be heated later to release the hydrogen.

Answers

Answer: A. Liquefy hydrogen under pressure and store it much as we do with liquefied natural gas today.

Explanation:

Current Hydrogen storage methods fall into one of two technologies;

physical storage where compressed hydrogen gas is stored under pressure or as a liquid; and chemical storage, where the hydrogen is bonded with another material to form a hydride and released through a chemical reaction.

Physical storage solutions are commonly used technologies but are problematic when looking at using hydrogen to fuel vehicles. Compressed hydrogen gas needs to be stored under high pressure and  requires large and heavy tanks. Also, liquid hydrogen boils at -253°C (-423°F) so it needs to be stored cryogenically with heavy insulation and actually contains less hydrogen compared with the same volume of gasoline.  

Chemical storage methods allow hydrogen to be stored at much lower pressures and offer high storage performance due to the strong binding of hydrogen and the high storage densities. They also occupy relatively smaller spaces than either compressed hydrogen gas or liquified hydrogen. A large number of chemical storage systems are under investigation, which involve hydrolysis reactions, hydrogenation/dehydrogenation reactions, ammonia borane and other boron hydrides, ammonia, and alane etc.

Other practical storage methods being researched that focuses on storing hydrogen as a lightweight, compact energy carrier for mobile applications include;

Metal hydrides  e.g. LiHNanostructured metal hydrides Non-metal hydrides Carbohydrates Synthesized hydrocarbons Aluminum Liquid organic hydrogen carriers (LOHC) Encapsulation , e.t.c.

1.When HCl(aq) and NaOH(aq) are mixed in a beaker, the beaker feels warm to the touch. What is known about the enthalpy of this reaction?a. The reaction is exothermic.b. The reaction is endothermic.c. Heat is absorbed from the surroundings.d. \DeltaH is positive.Part B 2. Consider this combination reaction: 2Mg(s)+O2(g)→2MgO(s) ΔH=−1204 kJ What is the enthalpy for the decomposition of 1 mole of MgO(s) into Mg(s) and O2(g)? View Available Hint(s) A. -602 kJ/mol B. 602 kJ/mol C. -1204 kJ/mol D. 1204 kJ/mol Part C 3. The enthalpy for the formation of 1 mole of NH3(aq) is -80.29 kJ/mol. What is the enthapy for the formation of 3 moles of NH3(aq)? View Available Hint(s) The enthalpy for the formation of 1 mole of is -80.29 . What is the enthapy for the formation of 3 moles of ? A -83.29 kJ B. -518×103 kJ C. -240.87 kJ D. -26.76 kJ

Answers

Answer:

1. The correct answer is option a.

2. The correct answer is option B.

3. The correct answer is option C.

Explanation:

1. When acid reacts with base heat is generated along with formation of salt and water.

[tex]HCl+NaOH\rightarrow NaCl+H_2O,\Delta H=Negative[/tex]

Those reaction in which heat released as a product is called exothermic reaction.Exothermic reaction have negative value of enthalpy of reaction

2. [tex]2Mg(s)+O_2(g)→2MgO(s), \Delta H=-1204 kJ[/tex]

If we reveres the equation we  will have the reaction in which MgO is getting decomposed into Mg and oxygen gas.

[tex]2MgO\rightarrow 2Mg+O_2(g),\Delta H=1204 kJ[/tex]

Divide the whole equation by 2.

[tex]MgO\rightarrow Mg+\frac{1}{2}O_2(g),\Delta H=602 kJ[/tex]

602 kJ is the enthalpy for the decomposition of 1 mole of MgO(s).

3.

The enthalpy for the formation of 1 mole of liquid ammonia = -80.29 kJ/mol

So, enthalpy of formation of 3 moles of  liquid ammonia :

[tex]3 mol\times (-80.29 kJ/mol)=-240.87 kJ[/tex]

-240.87 kJ is the enthapy for the formation of 3 moles of liquid ammonia.

The ozonization of pentene in carbon tetrachloride solution at 25 o C5H10 + O3 C5H10O3 is first order in C5H10 and first order in O3.

1.Complete the rate law for this reaction in the box below.
Use the form k[A]m[B]n... , where '1' is understood for m, n ...(don't enter 1) and concentrations taken to the zero power do not appear.

Rate =

2.In an experiment to determine the rate law, the rate of the reaction was determined to be 649 Ms-1 when [C5H10] = 0.128 M and [O3] = 4.41E-2 M. From this experiment, the rate constant is ______M-1s-1.

Answers

1. The rate law is Rate = k[C5H10][O3].

2. The rate constant is 1.24E6 M-1s-1.

1. The rate law for the ozonization of pentene in carbon tetrachloride solution at 25 o C5H10 + O3 C5H10O3 is first order in C5H10 and first order in O3. Therefore, the rate law can be written as:

Rate = k[C5H10][O3]

2. The rate constant can be calculated using the following equation:

k = Rate / [C5H10][O3]

Substituting the known values into the equation above, we get the following:

k = 649 Ms-1 / 0.128 M * 4.41E-2 M

k = 1.24E6 M-1s-1

Therefore, the rate constant for the ozonization of pentene in carbon tetrachloride solution at 25 o C5H10 + O3 C5H10O3 is 1.24E6 M-1s-1.

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What is the Molecule Geometry of a ABE3 molecule?

A. linear
B. trigonal bipyramidal
C. trigonal planar
D. tetrahedral

Answers

Final answer:

The Molecule Geometry of an ABE3 molecule, with three electron groups and no lone pairs, is trigonal planar, with groups arranged 120° apart in a plane.

Explanation:

The Molecule Geometry of a molecule with the designation ABE3, where 'A' represents the central atom and 'E' represents electron groups (or bonding domains) surrounding the central atom, relates to how these groups are spatially distributed around the central atom. In an ABE3 molecule configuration, with three electron groups around the central atom and no lone pairs (indicated by the '3'), all three groups are arranged to be as far apart from one another as possible. This arrangement forms trigonal planar geometry, where the groups adopt the positions at the corners of an equilateral triangle, each 120° apart and all in the same plane.

1. The most useful ore of aluminum is bauxite, in which Al is present as hydrated oxides, Al2O3⋅xH2O The number of kilowatt-hours of electricity required to produce 4.00kg of aluminum from electrolysis of compounds from bauxite is ________ when the applied emf is 5.00V. 2. The most useful ore of aluminum is bauxite, in which Al is present as hydrated oxides, The number of kilowatt-hours of electricity required to produce of aluminum from electrolysis of compounds from bauxite is ________ when the applied emf is a. 59.6 b. 0.0168 c. 0.0596 d. 19.9 e. 39.7

Answers

*Answer:

Option A: 59.6

Explanation:

Step 1: Data given

Mass of aluminium = 4.00 kg

The applied emf = 5.00 V

watts = volts * amperes

Step 2: Calculate amperes

equivalent mass of aluminum = 27 / 3 = 9  

mass of deposit = (equivalent mass x amperes x seconds) / 96500

4000 grams = (9* amperes * seconds) / 96500

amperes * seconds = 42888888.9

1 hour = 3600 seconds

amperes * hours = 42888888.9 / 3600 = 11913.6

amperes = 11913.6 / hours

Step 3: Calculate kilowatts

watts = 5 * 11913.6 / hours

watts = 59568 (per hour)

kilowatts = 59.6 (per hour)

The number of kilowatt-hours of electricity required to produce 4.00kg of aluminum from electrolysis of compounds from bauxite is 59.6 kWh when the applied emf is 5.00V

A voltaic cell is constructed in which the anode is a Zn|Zn2+ half cell and the cathode is a Fe2+|Fe3+ half cell. The half-cell compartments are connected by a salt bridge.(Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.)The anode reaction is: ____+_____ yeilds _____ +______The cathode reaction is: _____+_______ yeilds_____ +________The net cell reaction is: _____+ _______yeilds _______+______In the external circuit, electrons migrate _______the Fe2+|Fe3+ electrode _______the Zn|Zn2+ electrode.In the salt bridge, anions migrate ________ the Fe2+|Fe3+ compartment ________ the Zn|Zn2+ compartment.

Answers

Final answer:

In a voltaic cell set up with Zn and Fe half cells, the anode reaction is Zn (s) → Zn2+ (aq) + 2e-, the cathode reaction is Fe3+ (aq) + e- → Fe2+ (aq), and the net cell reaction is Zn (s) + Fe3+ (aq) → Zn2+ (aq) + Fe2+ (aq). Electrons move from the Zn electrode to the Fe electrode and anions in the salt bridge migrate from the Zn compartment to the Fe compartment.

Explanation:

In a voltaic cell constructed in which the anode is a Zn|Zn2+ and the cathode is a Fe2+|Fe3+ half cell, the reactions would look something like this:

Anode reaction: Zn (s) → Zn2+ (aq) + 2e-

Cathode reaction: Fe3+ (aq) + e- → Fe2+ (aq)

Net cell reaction: Zn (s) + Fe3+ (aq) → Zn2+ (aq) + Fe2+ (aq)

In the external circuit, electrons migrate from the Zn|Zn2+ electrode to the Fe2+|Fe3+ electrode. In the salt bridge, anions migrate from the Zn|Zn2+ compartment to the Fe2+|Fe3+ compartment.

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The solubility of oxygen in lakes high in the Rocky Mountains is affected by the altitude. If the solubility of O 2 from the air is 2.67 × 10 –4 M at sea level and 25°C, what is the solubility of O 2 at an elevation of 12,000 ft where the atmospheric pressure is 0.657 atm?

Assume the temperature is 25°C, and that the mole fraction of O 2 in air is 0.209 at both 12,000 ft and at sea level. 1.75 × 10–4 M 2.67 × 10–4 M 3.66 × 10–5 M 4.06 × 10–4 M

Answers

Answer:

S = 1,75x10⁻⁴M

Explanation:

It is possible to answer this question using Henry's law that states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid. The formula is:

S = k×Pi

Where S is solubility, K is Henry's constant and Pi is partial pressure.

At sea level, Pi of O₂ is:

1atm×0,209 = 0,209 atm. Replacing:

2,67x10⁻⁴M = k×0,209atm

k = 1,28x10⁻³M/atm

At 12,000ft Pi of O₂ is:

0,657atm×0,209 = 0,137 atm. Replacing:

S = 1,28x10⁻³M/atm×0,137atm

S = 1,75x10⁻⁴M

I hope it helps!

I have a cup of hot coffee at 140 oC but I want to cool it to 110 oC. My cup holds about 0.3 kg of coffee. Fortunately, I have a bunch of aluminum cubes in the freezer that I can drop into my hot coffee to cool it down.

If each aluminum cube has a mass of 1 g (not 1 kg!) and my freezer keeps its contents at a temperature of –10 oC, how many cubes do I have to drop into my coffee? The specific heat of water is around 4000 joules/kg/oC and aluminum is about 900 joules/kg/oC. (Pick the answer closest to the true value and ignore any thermal losses to surroundings.)

A. 200
B. 330
C. 400
D. 110
E. 88

Answers

Answer:

The correct answer is option B.

Explanation:

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

[tex]q_1=-q_2[/tex]

[tex]m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)[/tex]

where,

[tex]C_1[/tex] = specific heat of metal = [tex]900 J/kg^oC[/tex]

[tex]C_2[/tex] = specific heat of coffee= [tex]4000 J/kg^oC[/tex]

[tex]m_1[/tex] = mass of metal = x

[tex]m_2[/tex] = mass of coffee = 0.3 kg

[tex]T_f[/tex] = final temperature of aluminum metal= [tex]110^oC[/tex]

[tex]T_1[/tex] = initial temperature of aluminum metal = [tex]-10^oC[/tex]

[tex]T_2[/tex] = initial temperature of coffee= [tex]140^oC[/tex]

Now put all the given values in the above formula, we get

[tex]x\times 900 J/kg^oC\times (110-(-10))^oC=-(0.3 kg\times 4000 J/kg^oC\times (110-140)^oC[/tex]

[tex]x=0.333 kg[/tex]

Mass of aluminum cubes = 0.3333 kg = 333.3 g

If mass of 1 cube is 1 gram, then numbers of cubes in 333.3 grams will be:

[tex]=\frac{333.3 g}{1 g}=333.3\approx 330[/tex]

330 cubes of aluminum cubes will be required.

An analytical chemist weighs out of an unknown diprotic acid into a volumetric flask and dilutes to the mark with distilled water. He then titrates this solution with solution. When the titration reaches the equivalence point, the chemist finds he has added of solution. Calculate the molar mass of the unknown acid. Be sure your answer has the correct number of significant digits.

Answers

Final answer:

The molar mass of an unknown diprotic acid is determined through a titration procedure. Using the volume and molarity of NaOH, we calculate the moles of NaOH, then divide by 2 to get the moles of diprotic acid. Dividing the mass of the acid by the moles gives us the molar mass.

Explanation:

In this problem, the student is asked to calculate the molar mass of an unknown diprotic acid. The task requires the understanding of acids, stoichiometry, and titration concepts in analytical chemistry.

A diprotic acid is an acid that can donate two protons, or hydrogen ions, per molecule in solution. As the question doesn't provide the exact values, I'll explain conceptually. The molar mass of the unknown acid is calculated by using the volume of NaOH needed to reach the equivalence point and the concentration of NaOH. We'd start by calculating the moles of NaOH used (moles = volume x molarity), then because it's a diprotic acid, for every mole of acid, two moles of NaOH are needed, so we'd divide the moles of NaOH by 2 to get moles of unknown acid (moles acid = moles NaOH / 2). Lastly to find the molar mass, divide the mass of the acid used by the moles of acid calculated. So, the molar mass = mass of acid / moles of acid.

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Final answer:

The titration analysis provides a method to calculate the molar mass of an unknown acid using the volume of a known solution required to reach the equivalence point of the reaction. The process involves incrementally adding the known solution to the unknown, and using the reaction stoichiometry to find the unknown concentration. The molar mass is then calculated from this concentration, the initial volume of solution, and the molecular formula of the reaction products.

Explanation:

Based on the information provided, we need further details to accurately calculate the molar mass of the unknown diprotic acid. However, we can understand the general process involved in calculating such values using titration data.

The process is generally as follows:

Titration analysis is conducted by incrementally adding a solution of known concentration (the titrant) to the unknown solution (analyte).The titrant and analyte undergo a chemical reaction of known stoichiometry. The volume of titrant required to reach the 'equivalence point' of the reaction - where both the titrant and analyte have been completely reacted - allows us to calculate the concentration of the unknown analyte solution. The molar mass of the unknown acid can then be calculated using the resulting concentration, the initial volume of solution, and the defined molecular formula and mass of the reaction products.

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Given these reactions, where X represents a generic metal or metalloid 1 ) H 2 ( g ) + 1 2 O 2 ( g ) ⟶ H 2 O ( g ) Δ H 1 = − 241.8 kJ 2 ) X ( s ) + 2 Cl 2 ( g ) ⟶ XCl 4 ( s ) Δ H 2 = + 361.7 kJ 3 ) 1 2 H 2 ( g ) + 1 2 Cl 2 ( g ) ⟶ HCl ( g ) Δ H 3 = − 92.3 kJ 4 ) X ( s ) + O 2 ( g ) ⟶ XO 2 ( s ) Δ H 4 = − 607.9 kJ 5 ) H 2 O ( g ) ⟶ H 2 O ( l ) Δ H 5 = − 44.0 kJ what is the enthalpy, Δ H , for this reaction? XCl 4 ( s ) + 2 H 2 O ( l ) ⟶ XO 2 ( s ) + 4 HCl ( g )

Answers

Answer:

-767,2kJ

Explanation:

It is possible to sum enthalpies of half-reactions to obtain the enthalpy of a global reaction using Hess's law. For the reactions:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁= −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂= +361.7 kJ

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl (g) ΔH₃= −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄= − 607.9 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅= − 44.0 kJ

The sum of (4) - (2) produce:

6) XCl₄(s) + O₂(g) ⟶ XO₂(s) + 2Cl₂(g) ΔH₆ = ΔH₄ - ΔH₂ = -969,6 kJ

(6) + 4×(3):

7) XCl₄(s) + 2H₂(g) + O₂(g) ⟶ XO₂(s) + 4HCl(g) ΔH₇ = ΔH₆ + 4ΔH₃= -1338,8 kJ

(7) - 2×(1):

8) XCl₄(s) + 2H₂O(g) ⟶ XO₂(s) + 4HCl(g) ΔH₈ = ΔH₇ - 2ΔH₁= -855,2kJ

(8) - 2×(5):

9) XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g) ΔH₉ = ΔH₈ - 2ΔH₅= -767,2kJ

I hope it helps!

Final answer:

To calculate the enthalpy change for the reaction XCl4 (s) + 2 H2O (l) ⟶ XO2 (s) + 4 HCl (g), we can use Hess's law and the enthalpy values of the given reactions.

Explanation:

The enthalpy change for the reaction XCl4 (s) + 2 H2O (l) ⟶ XO2 (s) + 4 HCl (g) can be calculated using Hess's law and the enthalpy values of the given reactions.

To represent the desired reaction, we can combine reactions 2, 3, 4, and 5 as follows:

X (s) + 2 Cl2 (g) ⟶ XCl4 (s) [multiply this reaction by 2] 1/2 H2 (g) + 1/2 Cl2 (g) ⟶ HCl (g) [multiply this reaction by 4] X (s) + O2 (g) ⟶ XO2 (s) [multiply this reaction by 2] H2 (g) + 1/2 O2 (g) ⟶ H2O (l)

By summing these equations, we get the desired equation:

XCl4 (s) + 2 H2O (l) ⟶ XO2 (s) + 4 HCl (g)

--

The enthalpy change for the reaction is ΔH = ΔH2 + ΔH3 + ΔH4 + ΔH5.

Write Lewis structural formulas for the following molecules:

NH3 SO2
CH3OH HNO2
N2 CH2O

Answers

Answer :  The Lewis-dot structure of the following molecules are shown below.

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

(1) The given molecule is, [tex]NH_3[/tex]

As we know that hydrogen has '1' valence electron, nitrogen has '5' valence electrons.

Therefore, the total number of valence electrons in [tex]NH_3[/tex] = 5 + 3(1)  = 8

According to Lewis-dot structure, there are 6 number of bonding electrons and 2 number of non-bonding electrons.

(2) The given molecule is, [tex]SO_2[/tex]

As we know that sulfur and oxygen has '6' valence electrons.

Therefore, the total number of valence electrons in [tex]SO_2[/tex] = 6 + 2(6)  = 18

According to Lewis-dot structure, there are 8 number of bonding electrons and 10 number of non-bonding electrons.

(3) The given molecule is, [tex]CH_3OH[/tex]

As we know that carbon has '4' valence electrons, hydrogen has '1' valence electron and oxygen has '6' valence electrons.

Therefore, the total number of valence electrons in [tex]CH_3OH[/tex] = 4 + 4(1) + 6  = 14

According to Lewis-dot structure, there are 10 number of bonding electrons and 4 number of non-bonding electrons.

(4) The given molecule is, [tex]HNO_2[/tex]

As we know that hydrogen has '1' valence electron, nitrogen has '5' valence electrons and oxygen has '6' valence electrons.

Therefore, the total number of valence electrons in [tex]HNO_2[/tex] = 1 + 5 + 2(6) = 18

According to Lewis-dot structure, there are 8 number of bonding electrons and 10 number of non-bonding electrons.

(5) The given molecule is, [tex]N_2[/tex]

As we know that nitrogen has '5' valence electrons.

Therefore, the total number of valence electrons in [tex]N_2[/tex] = 2(5) = 10

According to Lewis-dot structure, there are 6 number of bonding electrons and 4 number of non-bonding electrons.

(6) The given molecule is, [tex]CH_2O[/tex]

As we know that carbon has '4' valence electrons, hydrogen has '1' valence electron and oxygen has '6' valence electrons.

Therefore, the total number of valence electrons in [tex]CH_2O[/tex] = 4 + 2(1) + 6  = 12

According to Lewis-dot structure, there are 8 number of bonding electrons and 4 number of non-bonding electrons.

When a student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant solution increases from 21.0 °C to 27.5 °C. Calculate the enthalpy change for the reaction in kJ per mol of HCl, assuming that the calorimeter loses only a negligible quantity of heat. The total volume of the solution is 100 mL, its density is 1.0 g/mL, and its specific heat is 4.18 J/g*K.

Answers

Answer: 54.4 kJ/mol

Explanation:

First we have to calculate the moles of HCl and NaOH.

[tex]\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=1.0M\times 0.05=0.05mole[/tex]

[tex]\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=1.0\times 0.05L=0.05mole[/tex]

The balanced chemical reaction will be,

[tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.05 mole of HCl neutralizes by 0.05 mole of NaOH

Thus, the number of neutralized moles = 0.05 mole

Now we have to calculate the mass of water:

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = [tex]50ml+50ml=100ml[/tex]

[tex]\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 100ml=100g[/tex]

Now we have to calculate the heat absorbed during the reaction.

[tex]q=m\times c\times (T_{final}-T_{initial})[/tex]

where,

q = heat absorbed = ?

[tex]c[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]

m = mass of water = 100 g

[tex]T_{final}[/tex] = final temperature of water = [tex]27.5^0C[/tex]

[tex]T_{initial}[/tex] = initial temperature of metal = [tex]21.0^0C[/tex]

Now put all the given values in the above formula, we get:

[tex]q=100g\times 4.18J/g^oC\times (27.5-21.0)^0C[/tex]

[tex]q=2719.6J=2.72kJ[/tex]

Thus, the heat released during the neutralization = 2.72 KJ

Now we have to calculate the enthalpy of neutralization per mole of [tex]HCl[/tex]:

0.05 moles of [tex]HCl[/tex] releases heat = 2.72 KJ

1 mole of [tex]HCl[/tex] releases heat =[tex]\frac{2.72}{0.05}\times 1=54.4KJ[/tex]

Thus the enthalpy change for the reaction in kJ per mol of HCl is 54.4 kJ

Answer:

54.34 kJ/mol of HCl

Explanation:

The calorimeter is a device used to determine the heat that is lost or gained, by a reaction. When the temperature change without a phase change, the heat (Q) can be calculated by:

Q = m*c*ΔT

Where m is the mass of the solution, c is the specific heat of the solution, and ΔT is the temperature variation (final - initial). The mass of the solution is the density multiplied by the volume:

m = 1.0 g/mL * 100 mL = 100 g

The temperature variation in °C is equal to the temperature variation in K, thus:

Q = 100g * 4.18J/gK * (27.5 - 21.0)K

Q = 2717 J

Thus, the solution gained 2717 J of heat. The enthalpy is how much of this energy is inside the matter, thus, it is the heat divided by the number of moles of a substance.

The number of moles of HCl is the volume (50 mL = 0.05 L) multiplied by the concentration:

n = 0.05 L * 1.0 M = 0.05 mol

The enthalpy is the heat divided by the number of moles:

H = 2717/0.05

H = 54340 J/mol of HCl

H = 54.34 kJ/mol of HCl

Percent Ionization
Percent ionization for a weak acid (HA) is determined by the following formula:

Percent ionization=[HA] ionized[HA] initial×100%

For strong acids, ionization is nearly complete (100%) at most concentrations. However, for weak acids, the percent ionization changes significantly with concentration. The more diluted the acid is, the greater percent ionization.
A certain weak acid, HA, has a Ka value of 7.6×10−7.
Part A
Calculate the percent ionization of HA in a 0.10 M solution.
Express your answer as a percent using two significant figures.

%

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Part B
Calculate the percent ionization of HA in a 0.010 M solution.
Express your answer as a percent using two significant figures.

%

Answers

Answer:

(a) 0.26 % (b) 0.80 %

Explanation:

(a)

Given that:

[tex]K_{a}=7.6\times 10^{-7}[/tex]

Concentration = 0.10 M

Considering the ICE table for the dissociation of acid as:-

[tex]\begin{matrix}&HA&\rightleftharpoons &A^-&&H^+\\ At\ time, t = 0 &0.10&&0&&0\\At\ time, t=t_{eq}&-x&&+x&&+x\\ ----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:-&0.10-x&&x&&x\end{matrix}[/tex]

The expression for dissociation constant of acid is:

[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [ {A}^- \right ]}{[HA]}[/tex]

[tex]7.2\times 10^{-7}=\frac{x^2}{0.10-x}[/tex]

[tex]7.2\left(0.10-x\right)=10000000x^2[/tex]

Solving for x, we get:

x = 0.00026  M

Percentage ionization = [tex]\frac{0.00026}{0.10}\times 100=0.26 \%[/tex]

(b)

Concentration = 0.010 M

Considering the ICE table for the dissociation of acid as:-

[tex]\begin{matrix}&HA&\rightleftharpoons &A^-&&H^+\\ At\ time, t = 0 &0.010&&0&&0\\At\ time, t=t_{eq}&-x&&+x&&+x\\ ----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:-&0.010-x&&x&&x\end{matrix}[/tex]

The expression for dissociation constant of acid is:

[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [ {A}^- \right ]}{[HA]}[/tex]

[tex]7.2\times 10^{-7}=\frac{x^2}{0.010-x}[/tex]

[tex]7.2\left(0.010-x\right)=10000000x^2[/tex]

Solving for x, we get:

x = 0.00008  M

Percentage ionization = [tex]\frac{0.00008}{0.010}\times 100=0.80 \%[/tex]

What is the half-life of A? What will the pressure, initially 32.1 kPa, at

(a) 10 s,
(b) 10 min after initiation of the reaction?

Answers

Answer:

a) 32.09 kPa

b) 32.09 kPa

Explanation:

Given data:

rate constant [tex]= 3.56\times 10^{-7} s^{-1}[/tex]

initial pressure is = 32.1 kPa

half life of A is calculated as

[tex]t_{1/2} = \frac{ln 2}{k}[/tex]

[tex]t_{1/2} = \frac{ln 2}{3.56\time 10^{-7}}[/tex]

[tex]t_{1/2} = = 1.956 \times 10^6 s[/tex]

for calculating pressure we have follwing expression

[tex]ln p = ln P_o - kt[/tex]

[tex]P =P_o e^{-kt}[/tex]

a) [tex]P = 32.1 e^{-3.56\times 10^{-7} \times 10} = 32.09 kPa[/tex]

b) [tex]P = 32.1 e^{-3.56\times 10^{-7} \times 10\times 60} = 32.09 kPa[/tex]

Rank the following materials from highest BP to lowest BP: a. N2, NaF, NH3, Ni, HIb. Ag, H20, Ar, HBr, KC c. LiF, Fe, F2. HF, HCI

Answers

Explanation:

Metallic solids are defined as the solids in which atoms of metals are held together by metallic bonds. These bonds are much stronger than an ionic bond.

In metallic solids, the electrons are delocalized in nature. On the other hand, ionic solids have strong bonds due to the presence of opposite charges on its combining atoms.

Molecular solids are defined as the solids in which atoms are combined together through Vander waal forces. Hence, molecular solids have low boiling point.

(a) It is known that [tex]N_{2}[/tex] is a molecular solid, NaF is an ionic solid, in [tex]NH_{3}[/tex] there exists hydrogen bonding between the molecules, and Ni is a metallic solid. Hence, the given materials are placed according to highest B.P. to lowest B.P. as follows.

            Ni > NaF > [tex]NH_{3}[/tex] > HI > [tex]N_{2}[/tex]

(b)   Ag is a metallic solid, and KCl is an ionic solid. Oxygen atom is more electronegative than bromine atom. So, the boiling point of [tex]H_{2}O[/tex] is more than the boiling point of HBr. Whereas Ar is a noble gas and it has low boiling point.

Hence, the given materials are placed according to highest B.P. to lowest B.P. as follows.

             Ag > KCl > [tex]H_{2}O[/tex] > HBr > Ar

(c)    Fe is also a metallic solid and electronegativity of fluorine is more than the electronegativity of chlorine. Hence, HF has high boiling point than HCl. And, [tex]F_{2}[/tex] being a covalent compound has weak intermolecular forces. So, the boiling point of [tex]F_{2}[/tex] will be the least.

Therefore, the given materials are placed from highest B.P. to lowest B.P. as follows.

           Fe > LiF > HF > HCl > [tex]F_{2}[/tex]  

Final answer:

To rank the boiling points from highest to lowest, identify the type of bonding and intermolecular forces present in each substance: ionic, metallic, hydrogen bonding, dipole-dipole interactions, or London dispersion forces. Ionic and metallic bonds generally lead to higher boiling points than molecular compounds.

Explanation:

The student has asked a Chemistry question regarding the ranking of the boiling points (BP) of different substances. Boiling points can be compared using intermolecular forces: ionic bonds, hydrogen bonds, dipole-dipole interactions, and London dispersion forces. Ionic compounds like NaF typically have higher boiling points than covalent compounds due to the strong electrostatic force between ions.

Metallic bonds found in metals like Ni also exhibit high boiling points. Among covalent compounds, the presence of hydrogen bonding, as found in NH3, would lead to a higher boiling point than compounds that rely solely on dipole-dipole interactions (such as HF) or London dispersion forces (like N₂ and HBr).

Given this, we would rank the boiling points from highest to lowest as:

a. NaF > Ni > NH₃ > HI > N₂,

b. Ag > HBr > KC > H₂O > Ar

c. F₂ > Fe > LiF > HCI > HF

assuming the compounds are in pure form and standard atmospheric pressure is taken into account.

How many electrons are transferred in the following reaction?

(The reaction is unbalanced.)

Mg(s) + Al3+(aq) → Al(s) + Mg2+(aq)

Answers

Answer:

6

Explanation:

Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

[tex]X\rightarrow X^{n+}+ne^-[/tex]

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

[tex]X^{n+}+ne^-\rightarrow X[/tex]

For the given chemical reaction:

[tex]Mg(s)+Al^{3+}(aq.)\rightarrow Al(s)+Mg^{2+}(s)[/tex]

The half cell reactions for the above reaction follows:

Oxidation half reaction:  [tex]Mg\rightarrow Mg^{2+}+2e^-[/tex]

Reduction half reaction:  [tex]Al^{3+}+3e^-\rightarrow Al[/tex]

Magnesium is loosing 2 electrons to form the magnesium cation. Thus, it is getting oxidized. Aluminum anion is gaining 3 electrons to form Aluminum. Thus, it is getting reduced.

Thus, balancing the half-reactions as:-

Oxidation half reaction:  [tex]3Mg\rightarrow 3Mg^{2+}+6e^-[/tex]

Reduction half reaction:  [tex]2Al^{3+}+6e^-\rightarrow 2Al[/tex]

Thus, total number of electrons transferred = 6

Answer:

3Mg +2Al^3 ⇆ 3 Mg^2+ + 2Al

In this reaction 6 electrons are transferred

Explanation:

Step 1: The half reactions

Mg  -2e- ⇆ Mg^2+

Al^3+ +3e- ⇆ Al

Step 2: Balance both equations

3*(Mg  -2e- ⇆ Mg^2+)

2(Al^3+ +3e- ⇆ Al)

3Mg  -6e- ⇆ 3Mg^2+

2Al^3+ +6e- ⇆ 2Al

Step 3: The netto reaction

3Mg +2Al^3 ⇆ 3 Mg^2+ + 2Al

In this reaction 6 electrons are transferred

Phenolphthalein has a pKa of 9.7 and is colorless in its acid form and pink in its basic form. calculate [In-}/{HIn} for the following pH

4.9

2.1

7.8

11.5

Answers

Answer:

1.58x10⁻⁵

2.51x10⁻⁸

0.0126

63.10

Explanation:

Phenolphthalein acts like a weak acid, so in aqueous solution, it has an acid form HIn, and the conjugate base In-, and the pH of it can be calculated by the Handerson-Halsebach equation:

pH = pKa + log[In-]/[HIn]

pKa = -logKa, and Ka is the equilibrium constant of the dissociation of the acid. [X] is the concentrantion of X. Thus,

i) pH = 4.9

4.9 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = - 4.8

[In-]/[HIn] = [tex]10^{-4.8}[/tex]

[In-]/[HIn] = 1.58x10⁻⁵

ii) pH = 2.1

2.1 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -7.6

[In-]/[HIn] = [tex]10^{-7.6}[/tex]

[In-]/[HIn] = 2.51x10⁻⁸

iii) pH = 7.8

7.8 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -1.9

[In-]/[HIn] = [tex]10^{-1.9}[/tex]

[In-]/[HIn] = 0.0126

iv) pH = 11.5

11.5 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = 1.8

[In-]/[HIn] = [tex]10^{1.8}[/tex]

[In-]/[HIn] = 63.10

For which of the following processes will \DeltaΔS be negative?PbCl2(s) = Pb2+(aq) + 2 Cl-(aq)MgO(s) + CO2(g) = MgCO3(s)CO2(aq) = CO2(g)C5H12(l) + 8 O2(g) = 5 CO2(g) + 6 H2O(g)

Answers

Answer

a) [tex]PbCl_2(s)\rightarrow Pb^{2+}(aq)+2Cl^-(aq)[/tex]: [tex]\Delta S[/tex] = +ve

b) [tex]MgO(s)+CO_2(g)\rightarrow MgCO_3(s)[/tex]: [tex]\Delta S[/tex] = -ve

c) [tex]CO_2(aq)\rightarrow CO_2(g)[/tex]: [tex]\Delta S[/tex]= +ve

d)  [tex]C_5H_{12}(l)+8O_2(g)\rightarrow 5CO_2(g)+6H_2O(g)[/tex]: [tex]\Delta S[/tex] = +ve

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from  an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

a) [tex]PbCl_2(s)\rightarrow Pb^(2+)(aq)+2Cl^-(aq)[/tex]

As solid is moving to ions form , randomness increases and thus sign of [tex]\Delta S[/tex] is positive.

b)[tex]MgO(s)+CO_2(g)\rightarrow MgCO_3(s)[/tex]

As gaseous reactants are converted to solid products , randomness decreases and thus sign of [tex]\Delta S[/tex]  is negative.

c) [tex]CO_2(aq)\rightarrow CO_2(g)[/tex]

As liquid is changing to gas  randomness increases and thus sign of [tex]\Delta S[/tex] is positive.

d) [tex]C_5H_{12}(l)+8O_2(g)rightarrow 5CO_2(g)+6H_2O(g)[/tex]

As 8 moles of gaseous reactants are converted to 11 moles of gaseous products , randomness increases and thus sign of [tex]\Delta S[/tex] is positive.

What are the coefficients of the reactants and products in the balanced equation above? Remember to include H2O(l) and OH−(aq) in the blanks where appropriate. Your answer should have six terms. Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4,4,3). Include coefficients of 1, as required, for grading purposes.

Answers

The order of coefficients of the reactants and products in the balanced reaction is 1,3,6,1,3,3.

[tex]BrO_3^- (aq) + Sn^{2+} + H^+ \longrightarrow Br^- + Sn^{4+} + H_2O.[/tex]

What is the balanced chemical equation?

The chemical equation in which the number of atoms of reactants and products is equal on either side of the equation is known as a balanced chemical equation.

According to the law of conservation of mass, the total mass on the reactant side should be equal to the total mass on the product side in a chemical equation

The given unbalanced chemical equation is as follows:

[tex]BrO_3^- (aq) + Sn^{2+} + H^+ \longrightarrow Br^- + Sn^{4+} + H_2O.[/tex]

To balance this equation, the coefficients are placed as shown below:

[tex]BrO_3^- (aq) + 3Sn^{2+} +6 H^+ \longrightarrow Br^- + 3Sn^{4+} + 3H_2O.[/tex]

Therefore, the order of the coefficient of reactants and products in the balanced equation is (1,3,6,1,3,3).

Learn more about the balanced chemical equation, here:

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Your question was incomplete, most probably the complete question was,

Consider the reaction given as

BrO₃⁻ (aq) + Sn²⁺ + __ → Br⁻ + Sn⁴⁺ + __

What are the coefficients of the reactants and products in the balanced equation above? Remember to include H2O(l) and H+(aq) in the appropriate blanks. Your answer should have six terms. Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4,4,3). Include coefficients of 1, as required, for grading purposes.

Scientists plan to place detection devices outside Earth's atmosphere to detect radiations from a star. Which of the following best explains the type of detection device that the scientists would use.a. A radar, because high frequency microwaves are limited to the region outside Earth's atmosphereb. Special cameras, because high frequency UV rays are emitted by objects at very high temperaturesc. Special cameras, because high energy UV rays are emitted by objects having very high massd. A radar, because high energy microwaves are produced from very hot objects

Answers

Answer: the correct option is A (A radar, because high frequency microwaves are limited to the region outside Earth's atmosphere)

Explanation:

A radar can be used for detection of devices outside Earth's atmosphere to detect radiations from a star. A radar system consists of a transmitter producing electromagnetic waves in the radio or microwaves domain, a transmitting antenna, a receiving antenna (often the same antenna is used for transmitting and receiving) and a receiver and processor to determine properties of the object(s) example the stars

Answer:

A radar, because high frequency microwaves are limited to the region outside Earth's atmosphere

Explanation:

I took the test and its correct

Express the following numbers in scientific notation:

a. 60200 L
b. 4520 J
c. 0.00600 mg
d. 0.01023 km
e. 80.70 mL

Answers

Answer :

(a) The number in scientific notation is, [tex]6.0200\times 10^4L[/tex]

(b) The number in scientific notation is, [tex]4.520\times 10^3J[/tex]

(c) The number in scientific notation is, [tex]6.00\times 10^{-3}mg[/tex]

(d) The number in scientific notation is, [tex]1.023\times 10^{-2}km[/tex]

(e) The number in scientific notation is, [tex]8.070\times 10^1mL[/tex]

Explanation :

Scientific notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.

For example :

5000 is written as [tex]5.0\times 10^3[/tex]

889.9 is written as [tex]8.899\times 10^{-2}[/tex]

In this examples, 5000 and 889.9 are written in the standard notation and [tex]5.0\times 10^3[/tex]  and [tex]8.899\times 10^{-2}[/tex]  are written in the scientific notation.

[tex]8.89\times 10^{-2}[/tex]  this is written in the scientific notation and the standard notation of this number will be, 0.00889.

If the decimal is shifting to right side, the power of 10 is negative and if the decimal is shifting to left side, the power of 10 is positive.

(a) As we are given the 60200 L in standard notation.

Now converting this into scientific notation, we get:

[tex]\Rightarrow 60200L=6.0200\times 10^4L[/tex]

As, the decimal point is shifting to left side, thus the power of 10 is positive.

Hence, the correct answer is, [tex]6.0200\times 10^4L[/tex]

(b) As we are given the 4520 J in standard notation.

Now converting this into scientific notation, we get:

[tex]\Rightarrow 4520J=4.520\times 10^3J[/tex]

As, the decimal point is shifting to left side, thus the power of 10 is positive.

Hence, the correct answer is, [tex]4.520\times 10^3J[/tex]

(c) As we are given the 0.00600 mg in standard notation.

Now converting this into scientific notation, we get:

[tex]\Rightarrow 0.00600mg=6.00\times 10^{-3}mg[/tex]

As, the decimal point is shifting to right side, thus the power of 10 is negative.

Hence, the correct answer is, [tex]6.00\times 10^{-3}mg[/tex]

(d) As we are given the 0.01023 km in standard notation.

Now converting this into scientific notation, we get:

[tex]\Rightarrow 0.01023km=1.023\times 10^{-2}km[/tex]

As, the decimal point is shifting to right side, thus the power of 10 is negative.

Hence, the correct answer is, [tex]1.023\times 10^{-2}km[/tex]

(e) As we are given the 80.70 mL in standard notation.

Now converting this into scientific notation, we get:

[tex]\Rightarrow 80.70mL=8.070\times 10^1mL[/tex]

As, the decimal point is shifting to left side, thus the power of 10 is positive.

Hence, the correct answer is, [tex]8.070\times 10^1mL[/tex]

A lead–tin alloy of composition 30 wt% Sn–70 wt% Pb is slowly heated from a temperature of 150°C (300°F).(a) At what temperature does the first liquid phase form?(b) What is the composition of this liquid phase?(c) At what temperature does complete melting of the alloy occur?(d) What is the composition of the last solid remaining prior to complete melting?

Answers

Answer:

a) 231.9 °C

b) 100% Sn

c) 327.5 °C

d) 100% Pb

Explanation:

This is a mixture of two solids with different fusion point:

[tex]Tf_{Pb}=327.5 C[/tex]

[tex]Tf_{Sn}=231.9 C[/tex]

Given that Sn has a lower fusion temperature it will start to melt first at that temperature.

So the first liquid phase forms at 231.9 °C and because Pb starts melting at a higher temperature, that phase's composition will be 100% Sn.

The mixture will be completely melted when you are a the higher melting temperature of all components (in this case Pb), so it will all in liquid phase at 327.5 °C.

At that temperature all Sn was already in liquid state and, therefore, the last solid's composition will be 100% Pb.

Which of the slightly soluble salts below will be more soluble in acidic solution than in pure water?

a. Mg(OH)2
b. CaCO3
c. AgCl Mg(OH)2
d. only Mg(OH)2,
e. CaCO3, and
f. AgCl
g. CaCO3 only
h. both Mg(OH)2 and CaCO3

Answers

Answer:

h. both Mg(OH)₂ and CaCO₃

Explanation:

Let's consider the solution of Mg(OH)₂ according to the following equation:

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

In acidic solution, OH⁻ reacts with H⁺ to form H₂O.

OH⁻(aq) + H⁺(aq) ⇄ H₂O(l)

According to Le Chatelier's principle, since [OH⁻] decreases, the solution of Mg(OH)₂(s) shifts toward the right, increasing its solubility.

Let's consider the solution of CaCO₃ according to the following equation:

CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)

In acidic solution, CO₃²⁻ reacts with H⁺ to form HCO₃⁻.

CO₃²⁻(aq) + H⁺(aq) ⇄ HCO₃⁻(aq)

According to Le Chatelier's principle, since [CO₃²⁻] decreases, the solution of CaCO₃(s) shifts toward the right, increasing its solubility.

Let's consider the solution of AgCl according to the following equation:

AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)

Cl⁻ does not react with H⁺ because it comes from a strong acid (HCl). Therefore, the solubility of AgCl(s) is not affected by the pH.

Final answer:

Mg(OH)₂ and CaCO₃ are more soluble in acidic solutions due to reactions with H+ ions that remove the OH- and CO₃²⁻ from the solution, driving the dissolution equilibrium forward.

Explanation:

The question asks which of the slightly soluble salts listed will be more soluble in acidic solution than in pure water. Specifically, salts like Mg(OH)₂ (magnesium hydroxide) and CaCO₃ (calcium carbonate) will be more soluble in an acidic solution. This is because the acid in the solution will react with the anionic part of the salt, which in the case of Mg(OH)2 is OH- and for CaCO₃ is CO₃²⁻. For example, in an acidic solution, H+ ions will react with OH- to form water, which effectively removes OH- from the solution and drives the dissolution equilibrium forward, increasing the solubility of Mg(OH)₂. Similarly, H+ ions will react with CO₃²⁻ to form HCO₃- (bicarbonate) or even further to H₂CO₃ (carbonic acid), which are more soluble than the carbonate ion, hence increasing the solubility of CaCO₃.

As for AgCl (silver chloride), it will also be more soluble in acidic solution because the chloride ion is not basic, and it does not react with H+ to form a weaker acid. Therefore, the correct answers that are more soluble in acidic solution than in pure water are Mg(OH)₂ and CaCO₃.

The half life for the decay of carbon-14 is 5.73 x 10^3 years. Suppose the activity due to the radioactive decay of the carbon-14 in a tiny sample of an artifact made of wood from an archeological dig is measured to be 53.Bq. The activity in a similar-sized sample of fresh wood is measured to be 59.Bq.
1. Calculate the age of the artifact. Round your answer to 2 significant digits.

Answers

Final answer:

To find the age of the artifact using carbon-14 activity, we calculate the ratio of current activity to original activity, use the decay formula with the half-life of carbon-14, and solve for time, yielding an estimated age of approximately 888 years.

Explanation:

To calculate the age of the artifact based on its carbon-14 activity, we use the concept of radioactive decay and the half-life of carbon-14. The half-life is the time taken for half of the radioactive atoms in a sample to decay. In the case of carbon-14, its half-life is 5,730 years.

We start by finding the ratio of the current activity to the original activity.

Original activity (fresh wood) = 59 Bq

Current activity (artifact) = 53 Bq

Ratio (current/original) = 53 Bq / 59 Bq = 0.898

Next, we use the decay formula: N(t) = N0 × (1/2)(t/T), where:

N(t) is the remaining amount of substance after time t

N0 is the original amount of substance

T is the half-life of the substance

t is the time that has passed

Plugging in the values we know: 0.898 = (1/2)(t/5730)

To solve for t, we take the natural logarithm of both sides of the equation:

ln(0.898) = ln((1/2)(t/5730))
ln(0.898) = (t/5730) × ln(1/2)

After calculating:

t = 5730 × ln(0.898) / ln(1/2) = 5730 × -0.107 / -0.693
t ≈ 888 years (rounded to two significant digits)

Therefore, the age of the artifact is approximately 888 years old.

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