In a constant‑pressure calorimeter, 60.0 mL of 0.300 M Ba(OH)2 was added to 60.0 mL of 0.600 M HCl. The reaction caused the temperature of the solution to rise from 23.65 ∘C to 27.74 ∘C. If the solution has the same density and specific heat as water ( 1.00 g/mL and 4.184J/g⋅K,) respectively), what is Δ???? for this reaction (per mole H2O produced)? Assume that the total volume is the sum of the individual volumes.

Answer:

I will work only one of the listed equations ... you follow the given example for the remaining reactions. Thank you :-)

Rxn 1: Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s)

a) E(Pt⁺²/Fe°) = - 1.668v

b) Process is Non-spontaneous if E(cell) < 0

Explanation:

Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s) ⇔

Pt°(s)|Pt⁺²[0.057M]║Fe⁺²[0.006M]|Fe°(s)

As written, Pt° is shown undergoing oxidation with Fe⁺² undergoing reduction. Applying the reduction potentials to the analytical equations for E(cell) and ΔG(cell) gives E(Pt/Fe⁺²) < 0 and ΔG(Pt/Fe⁺²) > 0 which indicate a non-spontaneous process. The following supports this conclusion.

E°(Fe⁺²) = -0.44v

E°(Pt⁺²) = +1.20v

E°(Pt/Fe⁺²) =E°(Redn) - E°(Oxidn) =E°(Fe⁺²) - E°(Pt⁺²)

= -0.44v - (+1.20v) = - 1.64v

[Fe⁺²] = 0.0066M

[Pt⁺²] = 0.057M

n = electrons transferred = 2

E(nonstd) = E°(std) - (0.0592/n)logQ);

Q = [Pt⁺²]/[Fe⁺²]

= -1.64v - (0.0592/2)log[0.057M]/[0.006M]v = -1.668v

Also, if ΔG(cell) > 0 => indicates non-spontaneous process

ΔG(Pt/Fe⁺²) = - nFE = -(2)(96,500Coulombs)((-1.664v) > 0 Kj => nonspontaneous rxn. (1 Coulomb-volt = 1 Kilojoule)

The Nernst equation is used to obtain the cell potential under standard conditions from the cell potential under nonstandard conditions.

Pt(s) + Fe^2+(aq) -------> Pt^2+ (aq) + Fe(s)

We have the following information from the question;

[Fe2+] = 0.0066M,   [Pt2+] = 0.057M

The standard reaction potential is; 1.18 V - (-0.44V) = 1.62 V

Using Nernst equation;

Ecell = E°cell - 0.0592/n log Q

Where;

Q = [Pt2+] /[Fe2+] =  0.057M/ 0.0066M = 8.636

Ecell = 1.62 V - 0.0592/2 log(8.636)

Ecell = 1.59 V

Cu(s) + 2Ag+(aq) -------> Cu2+(aq) + 2Ag(s)

The standard reaction potential is; 0.80 V - 0.34 V = 0.46 V

Q = [Cu2+]/[Ag+]^2 = 0.013/0.013^2 = 76.92

Ecell =  0.46 V - 0.0592/2 log(76.92)

Ecell = 0.40 V

Co2+(aq) + Ti3+(aq) -------> Co3+(aq) + Ti2+(aq)

The standard reaction potential of the reaction is; 1.92 V  - (-0.37) = 2.29 V

Q = [Ti2+] [Co3+]/[Co2+] [Ti3+] = [0.0110] [0.030]/[0.050] [0.0055] = 0.00033/0.000275 = 1.2

Ecell = 2.29 V - 0.0592/1 log(1.2)

Ecell = 2.28 V

Mg(s) + Sn2+(aq) --------> Mg2+(aq) + Sn(s)

The standard reaction potential of the reaction is; (-0.14 V) - (-2.37 V) = 2.23 V

Q = [Mg^2+]/[Sn^2+] = [0.796 M]/[0.0170 M] = 46.8

Ecell = 2.23 V - 0.0592/2 log(46.8)

Ecell = 2.18 V

Each of the cell is spontaneous as written since Ecell in each case is positive.

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D: 3, 2, 3, 1 The Pb3(PO4) comes from having 3 Pb and 2 H3PO4. The H2 comes from the 2H3PO4

Answer: B got it correct

Explanation:

The  the mass of the cone using the triple beam balance will be 543.0 grams. correct option is A.

The triple beam balance is an instrument or apparatus  used to measure mass very accurately and  precisely and these  devices typically have a reading error or round off of  of ±0.05 grams.

The name of the instrument  refers to its three beams, in which  the middle beam is the largest and longest the far beam of medium size one average  and the front beam the smallest and lightest.

Therefore, the mass of the cone using the triple beam balance will be 543.0 grams. correct option is A.

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Answer:

A. CaCO₃(s) → CaO(s) + CO₂(g).

B. Decomposition reaction.

C. 29.46 g.

D. 93.94%

Explanation:

A. Write the balanced the reaction

CaCO₃(s) → CaO(s) + CO₂(g),

1.0 mol of CaCO₃ is decomposed to produce 1.0 mol of CaO and 1.0 mol of CO₂.

B. Identify the reaction type (single replacement, double replacement, synthesis, decomposition or combustion).

A decomposition reaction is a type of chemical reaction in which a single compound breaks down into two or more elements or new compounds.

Herein, CaCO₃ is decomposed to produce CaO and CO₂.

C. What is the theoretical yield of carbon dioxide gas (in grams)?

n = mass/molar mass = (67.0 g)/(100.08 g/mol) = 0.6695 mol.

Using cross-multiplication:

1.0 mol of CaCO₃ produces → 1.0 mol of CO₂, from stichiometry.

∴ 0.6695 mol of CaCO₃ produces → 0.6695 mol of CO₂.

∴ The theoretical yield of carbon dioxide gas = n*molar mass = (0.6695 mol)(44.0 g/mol) = 29.46 g.

D. What is the percent yield if 27.67 grams of carbon dioxide gas are actually produced?

∵ the percent yield = (actual yield/theoretical yield)*100.

actual yield = 27.67 g, theoretical yield = 29.46 g.

∴ the percent yield = (27.67 g/29.46 g)*100 = 93.94%.

The empirical formula of the hydrocarbon is C1H3, calculated from the products of combustion. Using the molecular weight approximation of 132 amu, the molecular formula determined is C8H24.

To determine the empirical formula of the hydrocarbon, we analyze the products from its combustion. Since the sample produced 0.4931 g of CO2, we calculate the moles of carbon:

0.4931 g CO2 × (1 mol CO2/44.01 g CO2) = 0.01120 mol CO2
0.01120 mol CO2 corresponds to 0.01120 mol C (since there is 1 C in each CO2 molecule).

Similarly, given 0.2691 g of H2O:

0.2691 g H2O × (1 mol H2O/18.02 g H2O) = 0.01493 mol H2O
This gives us 0.02986 mol H (since there are 2 H in H2O).

The molar ratio of C to H can be simplified by dividing by the smaller number of moles:

Molar ratio C:H = 0.01120 mol C : 0.02986 mol H
We reduce this ratio to the simplest whole numbers to get the empirical formula:

(0.01120 / 0.01120) : (0.02986 / 0.01120) = 1:2.67
Approximating to whole numbers, we get C1H3 as the empirical formula.

To find the molecular formula, we use the given molecular weight (approximately 132 amu). Since the empirical formula weight of C1H3 is 12 (for C) + 3 (for H) = 15 amu, we calculate the multiplier:

Molecular weight / Empirical formula weight = 132 amu / 15 amu ≈ 8.8
This multiplier indicates the molecular formula is approximately 8 or 9 times the empirical formula. A whole number will result when using 8, thus we get the molecular formula C8H24.

Answer:

Explanation:

The number of half-lives elapsed, n, is calculated dividing the time by the half-life time:

a. A sample of Ce-141 with a half-life of 32.5 days after 32.5 days

b. A sample of F-18 with a half-life of 110 min after 330 min

c. A sample of Au-198 with a half-life of 2.7 days after 5.4 days

Answer:

D) The nuclei of unstable isotopes  

Explanation:

For example, the carbon-14 in carbon dating is radioactive.

When it decays, a neutron in the nucleus is converted into a proton, an electron, and an antineutrino.

[tex]_{0}^{1}\text{n} \longrightarrow \, _{1}^{1}\text{p} + \, _{-1}^{0}\text{e} + \overline{\nu}_{\text{e}}[/tex]

When ln K is negative, it means the equilibrium constant K is less than one, which results in a positive Gibbs free energy (ΔGrxn), indicating that the reaction is spontaneous in the reverse direction.

If ln K is negative, this indicates that the equilibrium constant, K, is less than one. According to thermodynamic principles, this means that the Gibbs free energy (ΔGrxn) for the reaction under standard conditions is positive, and therefore, the reaction is spontaneous in the reverse direction.

This is because the natural logarithm of a number less than one yields a negative value, and since ΔG° is related to ln K by the negative product with the gas constant (R) and temperature (T), ΔG° becomes positive when ln K is negative. So, the correct statement is 'ΔGrxn is positive and the reaction is spontaneous in the reverse direction.' A negative ΔGrxn would indicate a reaction that is spontaneous in the forward direction. If ΔGrxn were zero, the system would be at equilibrium and K would equal one.

Answer:

An oxidation-reduction reaction  

Explanation:

[tex]\stackrel{\hbox{-2}}{\hbox{C}}\text{$_{6}$H$_{12}$O$_{6}$} + 6\stackrel{\hbox{0}}{\hbox{O}\text{$_{2}$}} \longrightarrow 6\stackrel{\hbox{+4}}{\hbox{C}}\text{O$_{2}$}} + 6\text{H$_{2}$}\stackrel{\hbox{-2}}{\hbox{O}}[/tex]

Look at the oxidation numbers of the atoms.

Each carbon atom in glucose loses four electrons (oxidation), and each oxygen atom in O₂ gains two electrons (reduction).

Answer:

Explanation:

1) Data:

a) Solution of a dye:

b) Water of the pool, after mixing the dye:

2) Formulae:

a) Molarity: M = n / V in liter

b) Molar mass: MM = mass in grams / number of moles

3) Solution:

a) Number of moles of methylene blue in the prepared solution:

b) Number of moles of methylen blue in the pool = number of moles of methylen blue in the prepared solution

c) Volume of the pool:

And that is the answer, which has to be rounded to 3 significant figures, since the data are expressed with 3 significant figures.

Note: For all the effects, the 50.0 mL of the dye solution are neglictible in front of the volume of the pool.

Final answer:

To calculate the volume of water in an irregularly shaped pool using a known concentration of methylene blue dye, the dilution formula is applied with the dye's initial concentration and mass, resulting in the pool's volume being 83,689 liters.

Explanation:

To find the volume of water in the pool using the concentration of methylene blue dye, first, determine the amount of dye initially used to disperse in the pool water. Since 1.00 g of methylene blue was dissolved in 50.0 mL of water, and knowing the molar mass of methylene blue is 319.854 g/mol, we can calculate the molarity of the initial solution before it was diluted in the pool.

Moles of methylene blue = mass (g) / molar mass (g/mol) = 1.00 g / 319.854 g/mol = 0.00313 moles.

Thus, the initial concentration of the dye in the solution dispersed in the pool is 0.00313 moles / 0.050 L = 0.0626 M.

When this dye disperses evenly throughout the pool, its concentration decreases to 3.74 × 10-8 M. The total volume of the pool water can be calculated by applying the concept of dilution, where the moles of dye remain constant but the volume changes:

Therefore, the volume of water in the pool is 83,689 liters.

The behavior of the cans of soft drinks in a malfunctioning refrigerator is explained by the properties of water's expansion on freezing and gas solubility changes under varying pressure and temperature, but none of the given options accurately explains why only diet soft drink cans ruptured.

The scenario with cans of soft drinks in a malfunctioning refrigerator can be explained by the behavior of carbonated beverages when exposed to changes in temperature and pressure. The key statements that describe this behavior are:

Therefore, while 'B. Water expands on freezing' is a correct statement, it does not fully explain why only diet soft drink cans ruptured. The differences in freezing point depression due to differing ingredient compositions between diet and regular soft drinks could provide a better explanation, albeit not directly addressed among the options given.

Answer:

1.135 M.

Explanation:

The reaction is a second order reaction of HI,so the rate law of the reaction is: Rate = k[HI]².

1/[A] = kt + 1/[A₀],

where, k is the reate constant of the reaction (k = 1.57 x 10⁻⁵ M⁻¹s⁻¹),

t is the time of the reaction (t = 8 hours x 60 x 60 = 28800 s),

[A₀] is the initial concentration of HI ([A₀] = ?? M).

[A] is the remaining concentration of HI after hours ([A₀] = 0.75 M).

∵ 1/[A] = kt + 1/[A₀],

∴ 1/[A₀] = 1/[A] - kt

∴ 1/[A₀] = [1/(0.75 M)] - (1.57 x 10⁻⁵ M⁻¹s⁻¹)(28800 s) = 1.333 M⁻¹ - 0.4522 M⁻¹ = 0.8808 M⁻¹.

∴ [A₀] = 1/(0.0.8808 M⁻¹) = 1.135 M.

So, the concentration of HI 8 hours earlier = 1.135 M.

Final answer:

To find the concentration of HI 8 hours earlier in a second-order reaction, we utilize the integrated rate law specific for second-order reactions and solve for the initial concentration using the given rate constant and the time in seconds.

Explanation:

The reaction 2HI → H2 + I2 is second order in [HI] and second order overall, meaning the rate equation can be written as Rate = k[HI]2. Given the rate constant (k) of 1.57 × 10−5 M−1s−1 at 700°C and the current concentration of 0.75 M, we need to calculate the concentration 8 hours earlier. To solve this, we can use the integrated rate law for a second-order reaction:

[HI]−1 - [HI]0−1 = kt

Where [HI]0 is the initial concentration, [HI] is the concentration after time t, k is the rate constant, and t is the time in seconds. We have:

[HI] = 0.75 M

k = 1.57 × 10−5 M−1s−1

t = 8 hours × 3600 seconds/hour = 28800 seconds

Plugging these values into the integrated rate equation, we calculate [HI]0. Finally, we can find the concentration 8 hours earlier.

Answer:

[tex]\boxed{\text{n + A $\longrightarrow$ B + C + n}}[/tex]

Explanation:

When a neutron hits the nucleus of a heavy atom, the nucleus splits into two fragments of roughly equal mass and emits two or three neutrons.

A model that fits the process is

[tex]\boxed{\textbf{n + A $\longrightarrow$ B + C + n}}[/tex]

Answer : The correct option is, [tex]A\rightarrow B+C+n[/tex]

Explanation :

Nuclear fusion : It is a process where the two small nuclei converted to form a heavy nuclei and release some energy.

Nuclear fission : It is a process where a heavier nuclei (unstable nuclei) converted into two or more small (stable nuclei) and release some energy.

From the given reactions, we conclude that the reaction that shows the nuclear fission reaction is:

[tex]A\rightarrow B+C+n[/tex]

Hence, the nuclear fission reaction is, [tex]A\rightarrow B+C+n[/tex]

Answer:

For a substance to classify as a mineral, it must lie within certain parameters. It should be an inorganic solid, that is naturally occurring in nature (not synthesized), with an ordered internal structure and a definite chemical composition.

By definite chemical composition, geologists mean that the mineral must be have chemical constituents that have an unvarying chemical composition, or a chemical composition that oscillates withing a very limited and specific range.

An example is the mineral, halite. It has a chemical composition of one sodium atom and one chloride atom, represented as NaCl and is unchanging in this composition throughout nature.

Answer:

The reaction will move to the left.

Explanation:

Ba(OH)₂ = Ba²⁺ + 2OH⁻,

Ba(OH)₂ is dissociated to Ba²⁺ and 2OH⁻.

H⁺ will combine with OH⁻ to form water.

So, the concentration of OH⁻ will decrease and the equilibrium is disturbed.

According to Le Châtelier's principle: when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.

So, the right choice is: the reaction will move to the left, is the choice that will not happen to the equilibrium.

Answer:

D.) The reaction will move to the left.

Explanation:

Answer:

Explanation:

1) Data:

a) Mass of CuSO₄ ∙ XH₂O = 1.50 g

b) Mass of CuSO₄ = 0.96

c) X = ?

2) Additional needed data:

a) Molar Mass of CuSO₄ = 159,609 g/mol

b) Molar mass of H₂O = 18,01528 g/mol

3) Chemical principles and formulae used:

a) Law of conservation of mass

b) Molar mass = mass in grams / number of moles = m / n

4) Solution:

a) Law of conservation of mass:

b) Moles

c) Proportion:

Divide both mole amounts by the least of the two numbers, i.e. 0.0060

Then, the ratio of CuSO₄ to H₂O is 1 : 5 and the chemical formula is:

Hence, the value of X is 5.

Answer:

CuSO4*5H2O

X = 5

Explanation:

Step 1: Data given

Mass of the copper(II)sulfate hydrate = 1.50 grams

After dehydration they find that they are left with 0.96 g of the an-hydrate CuSO4.

Step 2: Calculate mass of water

Mass of water = mass of hydrate - mass of anhydrate

Mass water = 1.50 grams - 0.96 grams

Mass water = 0.54 grams

Step 3: Calculate moles of water

Moles H2O = mass / molar mass

Moles H2O = 0.54 grams / 18.02 g/mol

Moles H2O = 0.030 moles

Step 4: Calculate moles of CuSO4

Moles CuSO4 = 0.96 grams / 159.61 g/mol

Moles CuSO4 = 0.0060 moles

Step 5: Calculate mol ratio

We divide by the smallest amount of moles

H2O : 0.030 / 0.060 = 5

CuSO4: 0.0060 / 0.0060 = 1

For 1 mol CuSO4 we have 5 moles of H2O

CuSO4*5H2O

X = 5

Answer:

7,72Е24

Explanation:

if one mole is 6.022*10²³, then 6.41 moles are: 6.41*6.022*10²³*2=2*3.86*10²⁴=7.72E24 (atoms).

PS/ Formula of Br is Br₂.

Answer:

Explanation:

1) Data:

a) Cylinder diameter: d = 17.0 cm

b) Cylinder height: H = 20.4 cm

c) p = 2.20 MPa

d) compound: BF₃ gas

e) mass, m = 0.0985 kg

2) Formulae:

a) Volume of a cylinder, V = π r² H

b) Number of moles, n = mass in grams / molar mass

c) Ideal gas equation: pV = nRT

3) Solution:

a) Volume (V)

i) r = d / 2 = 17.0 cm / 2 = 8.50 cm

ii) V = V = π r² H = π (8.50 cm) ² (20.4 cm) = 4,630 cm³ = 4.630 liter

b) Number of moles (n)

i) molar mass BF₃: 67.82 g/mol

ii) n = mass in grams / molar mass = 98.5 g / 67.82 g/mol = 1.452 mol

c) Temperature

i) pressure conversion: 2.20 MPa = 22.21 atm

ii) pV = nRT ⇒ T = pV / (nR)

The result is given with 3 significant figures, since that is the number of significant figures used for the data.

To calculate the maximum safe operating temperature for the gas reaction vessel, use the ideal gas law to find the number of moles of boron trifluoride gas and then calculate the maximum safe operating temperature.

To calculate the maximum safe operating temperature for the gas reaction vessel, we need to consider the ideal gas law and the given quantities of gas. First, we convert the pressure from megapascals to pascals. Then, we use the ideal gas law equation to find the number of moles of boron trifluoride gas. Finally, we use the molar mass of boron trifluoride to convert the number of moles to grams, and then divide by the volume of the vessel to find the density. By rearranging the ideal gas law equation to solve for temperature, we can calculate the maximum safe operating temperature.

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Organic molecules cause the least increase in conductivity when added to pure water, as they do not dissociate into ions, unlike ionic compounds which dissociate completely, and weak acids and bases which partially ionize.

The addition of organic molecules to pure water causes the least increase in conductivity. This is because organic molecules typically do not dissociate into ions when dissolved in water. On the other hand, ionic compounds dissociate almost completely in water, making them strong electrolytes and significantly increasing water's conductivity. Option D is correct .

Weak bases and acetic acid, a weak acid, only partially ionize in water, which would lead to a slight increase in conductivity compared to organic molecules, but much less than that caused by ionic compounds.

The weak bases mentioned do not refer to the dissociation of ionic compounds themselves but to the reactions of their polyatomic anions with water. Thus, when discussing weak bases in this context, it is important to consider these secondary reactions rather than primary dissociation.

Furthermore, acetic acid, being a weak acid, increases the hydronium ion concentration in water, but again not as much as a strong acid would. However, since both acetic acid and weak bases only partially ionize, they’re still considered weak electrolytes.

Answer: K

Hope this helps you out!

Answer:

Potassium (K)

Explanation:

The reason that potassium is the more reactive metal is because it is in group 1, as opposed to group 11 for copper, and being in group 1 means that it has one valence electron, which it can give away by reacting with almost any other element, and in doing so, it becomes much more stable.

The answer is C). Good luck.

If the solubility of potassium nitrate (KNO3) is 320 g/dm3,  potassium nitrate can be dissolved in 16 cm3 of water is 5.12 g. The correct option is C.

The solubility is the ability of a solute to gets dissolved in the solvent and form a solution.

The solubility for (KNO3) is 320 g/dm3.

Volume of water to be dissolved is

16cm³ = 0.016 dm³

Then the solubility for the given volume will be

320 g/dm³ x 0.016 dm³ = 5.12 g

Thus, if the solubility of potassium nitrate (KNO3) is 320 g/dm3,  potassium nitrate can be dissolved in 16 cm3 of water is 5.12 g. The correct option is C.

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Answer:

b) Gain or lose electrons

Explanation:

An ion is an electrically charged particle. For an atom to be charged, it must have gained or lost electron in the process and therefore, it becomes an ion.

The loss or gain of electrons is what makes an atom charged and eventually becomes an ion.

A positively charged ion is one that has lost an electron and it is called a cation. In such an ion, the number of electrons are lesser than those of protons. This is why they are cations

A negatively charged ion is one that has gained electrons. They are called anions. In such an ion, the number of electrons are greater than that of protons.

Answer:

D

Explanation:

What an interesting amusing question.

The formula for KE = 1/2 m v^2.

v implies that there is motion.

The answer you are looking for is the body of water that has 0 for v.

A cloud moves, so it is not the answer.

A hailstone moves until it hits the ground. I'm assuming it's either on it's way down or it is caught in some sort of updraft in a cumulonimbus cloud. So it has motion.

Same comment for a raindrop. It has a v until it hits the ground so I'm assuming it is not the answer.

I've never seen a puddle of water move. It can only evaporate. So of all your choices, this one (D) is likely the right answer.

Answer:

If the planet has a lower orbital radius it means the planets proximity to the sun is farther away. The climate of the planet will be colder than a planet that would have a closer proximity to the sun.

Answer:

Explanation:

1) Word equation (given):

• Dinitrogen gas + dihydrogen gas → gaseous ammonia.

2) Chemical equation:

• N₂ (g) + 3H₂ (g) → 2NH₃ (g)

3) Mole ratios:

• 1 mole N₂ (g) : 3 mole H₂ (g) : 2 mole NH₃ (g)

4) Reaction rates:

• Rate of dinitrogen consumption: r₁ = n₁ / t (moles/s)

• Rate of ammonia production: r₂ = n₂ / t (moles/s)

• Due to the stoichiometric ratios: r₂ = 2 × r₁

5) Calculate r₁:

• 284 liter / s

• PV = nRT ⇒ n = PV / (RT)

• Divide by time, t: n/t = P (V/t) / (RT)

• Substitute V/t = 284 liter/s, P = 0.75 atm, and T = 196 +273.15K = 469.15K

r₁ = n₁ / t = (0.75 atm) (284 liter/s) / [ (0.08206 atm-liter/K-mol) (469.15k) ]

= 5.53 moles/s

6) Calcualte r₂

7) Convert rate in mole/s to rate in kg/s

When the reaction is run at 196.°C and 0.75atm, the rate at which ammonia is being produced is 0.19 kg/s.

To calculate the rate:

Firstly mole ratios:

1 mole N₂ (g) : 3 mole H₂ (g) : 2 mole NH₃ (g)

The rate of dinitrogen consumption:

Rate of ammonia production: r₂ = n₂ / t (moles/s)

The stoichiometric ratios: r₂ = 2 × r₁

To calculate r₁:

284 liter / s

PV = nRT

n = PV / (RT)

t: n/t = P (V/t) / (RT)

V/t = 284 liter/s,

P = 0.75 atm,

T = 196 +273.15K = 469.15K

r₁ = n₁ / t = (0.75 atm) (284 liter/s) / [ (0.08206 atm-liter/K-mol) (469.15k) ]

= 5.53 moles/s

Find r₂

r₂ = 2 × r₁

= 2 × 5.53 mole/s

= 11.06 mole/s

7) Change rate in mole/s to rate in kg/s

mass in grams = molar mass × number of moles

molar mass of NH₃ = 17.03 g/mol

mass = 17.03 g/mol × 11.06 /s

= 188.4 g/s

Convert 188.4 g/s into kg/s: 0.1884 kg/s

The significant digits is 0.19 kg/s.

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Answer:

Explanation:

1) Ionization equilibrium equation: given

2) Ionization equilibrium constant, at 25°C, Kw: given

3) Stoichiometric mole ratio:

As from the ionization equilibrium equation, as from the fact it is stated, the concentration of both ions, at 25°C, are equal:

4) A solution has a [OH⁻] = 3.4 × 10⁻⁵ M at 25 °C and you need to calculate what the [H₃O⁺(aq)] is.

Since the temperature is 25°, yet the value of Kw is the same, andy you can use these conditions:

Then you can substitute the known values and solve for the unknown:

As you see, the increase in the molar concentration of the ion [OH⁻] has caused the decrease in the molar concentration of the ion [H₃O⁺], to keep the equilibrium law valid.

Answer:

1. pH = -log[H^+]

2. Acidic: pH < 7.00, neutral: pH = 7.00, basic: pH > 7.00

3. pH + pOH = 14.00 at room temperature

Explanation:

Answer:

Explanation:

Pair  2.50g of O₂ and 2.50g of  N₂

The atoms sample with the largest number of moles since the masses are the same would be the one with lowest molar mass according the the equation below:

Number of moles = [tex]\frac{mass }{molarmass}[/tex]

Atomic mass of O = 16g and N = 14g

Molar mass of O₂ = 16 x 2 = 32gmol⁻¹

Molar mass of N₂ = 14 x 2 = 28gmol⁻¹

Number of moles of O₂ = [tex]\frac{2.5}{32}[/tex] = 0.078mole

Number of moles of N₂ = [tex]\frac{2.5}{28}[/tex] =  0.089mole

We see that N₂ has the largest number of moles

Answer:

Pair A : 2.50 g N2    Pair B : 21.5 g n2    Pair C : 0.081 CO2

Explanation:

Answer:

Si: 1s²2s²2p⁶3s²3p2

Explanation:

Other electron configuration summaries include ...

Si:[Ne]3s²3p2; [Ne] = electron configuration of noble gas Neon (1s²2s²2p⁶)

Si:[Ne]3s²2p₋₁¹p₀¹p₊₁⁰ <=> e⁻ configuration with orbital orientations

Si:[Ne]3s(↑↓)3p₋₁(↑)p₀(↑) <=> Orbital Diagram

Final answer:

The ground-state electron configuration for silicon (Si) is determined by the Aufbau principle, which gives us Si: 1s²2s²2p¶3s²3p².

Explanation:

The ground-state electron configuration for silicon (Si), which has an atomic number of 14, is determined using the Aufbau principle. Starting with the lowest energy subshell, we continue to fill each subshell according to the prescribed order until all 14 electrons have been placed. Thus, following the order: 1s, 2s, 2p, 3s, 3p, the electron configuration for silicon will be:

Si: 1s²2s²2p¶3s²3p².

Here, each number represents the principal quantum number corresponding to the shell, while the letter indicates the subshell type. The superscript denotes the number of electrons in that particular subshell.

Answer:

D. Zn → Zn²⁺ + 2e⁻, 2H⁺ + 2e⁻ → H₂.

Explanation:

Oxidation reaction:

Zn losses 2 electrons and is oxidized to Zn²⁺:

Zn → Zn²⁺ + 2e⁻.

Reduction reaction:

H⁺ gains 1 electron and is reduced to H:

2H⁺ + 2e⁻ → H₂.

So, the right choice is: D. Zn → Zn²⁺ + 2e⁻, 2H⁺ + 2e⁻ → H₂.

Answer:

D. Zn → Zn²⁺ + 2e⁻, 2H⁺ + 2e⁻ → H₂.

Explanation:

Answer via Educere/ Founder's Education