Examine the false statement. In the Bohr model of an atom, an electron's path is not known exactly. Which answer choice rewords the false statement so it is true?
a.In the Bohr model of an atom, an electron's path follows a defined orbit inside the nucleus.
b.In the Bohr model of an atom, an electron's path follows a defined orbit around the nucleus.
c.In the Bohr model of an atom, an electron's path follows an erratic pattern inside the nucleus.
d.In the Bohr model of an atom, an electron's path follows an erratic pattern around the nucleus.

The question is incomplete, here is a complete question.

Calculate the standard entropy of vaporization of ethanol at its boiling point 352 K. The standard molar enthalpy of vaporization of ethanol at its boiling point is 40.5 kJ/mol.

Answer : The standard entropy of vaporization of ethanol is, 115 J/mol.K

Explanation :

Formula used :

[tex]\Delta S=\frac{\Delta H_{vap}}{T_b}[/tex]

where,

[tex]\Delta S[/tex] = change in entropy

[tex]\Delta H_{vap}[/tex] = change in enthalpy of vaporization = 40.5 kJ/mol

[tex]T_b[/tex] = boiling point temperature = 352 K

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

[tex]\Delta S=\frac{\Delta H_{vap}}{T_b}[/tex]

[tex]\Delta S=\frac{40.5kJ/mol}{352K}[/tex]

[tex]\Delta S=\frac{40.5\times 10^3J/mol}{352K}[/tex]

[tex]\Delta S=115J/mol.K[/tex]

Therefore, the standard entropy of vaporization of ethanol is, 115 J/mol.K

Final answer:

Positive ions, which form the nucleus, are considered fixed during the electrons’ oscillations due to their significantly larger mass, which makes them relatively stationary compared to the lightweight and mobile electrons. In atomic models, this assumption simplifies the study of electronic behavior.

Explanation:

Positive ions can be considered to be fixed during the electrons’ oscillations because of their relatively large mass compared to electrons. In the context of atomic physics and the Bohr model, positive ions are essentially the nucleus of an atom, which is comprised of protons and neutrons. These particles are much heavier than the electrons and thus remain relatively stationary when the electrons oscillate or move in their orbits.

Within the atom, cations, which are positive ions, are created when elements lose one or more electrons. For example, group 1 elements in the periodic table lose one electron easily due to their electronic configuration, leading to a positive charge. The difference in mass means that while the electrons, which are lightweight and mobile, can oscillate or change their energy states quickly, the heavier protons in the nucleus (the cations) do not move significantly during these processes. Consequently, in many atomic models and explanations of electronic behavior, the positive ions are often treated as if they are fixed in place.

Answer: Christine Herman & L.G Wade Jr., "2010". Organic Chemistry: Reaction of Alkane, 7e, Pearson Education, Radford University, Radford, VA.

Explanation:

This is an edited book. The Harvard reference style was used in the following order:

Authors name

Year of publication

Title

Edition

Publisher

Place of publication.

Note that the title of book should be italicized with capitalization of first word.

Answer:

The answer to your question is below

Explanation:

There are 4 types of chemical reactions:

- Synthesis is when two elements or compounds form only one compound.

- Decomposition is when 1 compound is broken into 2 or more products.

- Single replacement is when one element is replaced by another element.

- Double replacement is when the cations of two compounds are interchanged.

1.- Decomposition                      2 Al₂O₃    ⇒    4 Al   +  3O₂

2.- Single replacement          Mg  +  2HNO₃   ⇒   Mg(NO₃)₂   +  H₂

3.- Combustion                      2C₆H₆  +  15O₂   ⇒   12CO₂   +   6H₂O

4.- Synthesis                          2Ag   +   S   ⇒   Ag₂S

5.- Double replacement      3Ca(OH)₂   + 2H₃PO₄   ⇒   Ca₃(PO₄)₂  + 6 H₂O

Answer:

The answer to your question is 15.7 amu

Explanation:

                          Abundance               Mass

Isotope 1                30%                         15

Isotope 2               70%                          16

Average atomic mass = (Abundance isotope 1 x abundance) +

                                       (Abundance isotope 2 x abundance)

Substitution

Average atomic mass =  (0.30 x 15) + (0.70 x 16)

Simplify

Average atomic mass = 4.5 + 11.2

Result

Average atomic mass = 15.7 amu

Answer:

28.16 °C

Explanation:

Considering that:-

Heat gain by water = Heat lost by metal

Thus,  

[tex]m_{water}\times C_{water}\times (T_f-T_i)=-m_{metal}\times C_{metal}\times (T_f-T_i)[/tex]

Where, negative sign signifies heat loss

Or,  

[tex]m_{water}\times C_{water}\times (T_f-T_i)=m_{metal}\times C_{metal}\times (T_i-T_f)[/tex]

For water:

Mass = 165 g

Initial temperature = 28 °C

Specific heat of water = 4.184 J/g°C

For metal:

Mass = 4.00 g

Initial temperature = 75 °C

Specific heat of water = 0.600 J/g°C

So,  

[tex]165\times 4.184\times (T_f-28)=4.00\times 0.600\times (75-T_f)[/tex]

[tex]690360\left(T_f-28\right)=2400\left(75-T_f\right)[/tex]

[tex]692760T_f=19510080[/tex]

[tex]T_f = 28.16\ ^0C[/tex]

Hence, the final temperature is 28.16 °C

Answer:

K = 3.37

Explanation:

2 NH₃(g) → N₂(g)  + 3H₂(g)

Initially we have 4 mol of ammonia, and in equilibrium we have 2 moles, so we have to think, that 2 moles have been reacted (4-2).

              2 NH₃(g)    →    N₂(g)  + 3H₂(g)

Initally       4moles             -            -

React        2moles           2m   +   3m

Eq             2 moles          2m        3m

We had produced 2 moles of nitrogen and 3 mol of H₂ (ratio is 2:3)

The expression for K is:  ( [H₂]³ . [N₂] ) / [NH₃]²

We have to divide the concentration /2L, cause we need MOLARITY to calculate K (mol/L)

K = ( (2m/2L) . (3m/2L)³ ) / (2m/2L)²

K = 27/8 / 1 → 3.37

Answer:

The value of K for this reaction is 1.69

Explanation:

Step 1: Data given

Moles of NH3 = 4.0 moles

Volume of the container = 2.0 L

At the equilibrium 2.0 moles NH3 remains

Step 2: The balanced equation

2 NH3(g) → N2(g) + 3H2(g)

Step 3: Initial number of moles

NH3: 4.0 moles

N2: 0 moles

H2: 0 moles

Step 4: Number of moles at the equilibrium

NH3: 2.0 moles

This means there reacts 2.0 moles of NH3

For 2 moles of NH3 we have 1 mol of N2 and 3 moles of H2

There will be produced 1 mol of N2 and 3 moles of H2

Step 5: Calculate molarity

Molarity = moles / volume

Molarity of NH3 = 2.0 moles / 2.0 L = 1 M

Molarity of N2 = 1.0 mol / 2.0 L = 0.5 M

Molarity of H2 = 3.0 mol / 2.0 L = 1.5 M

Kc = ([H2]³[N2]) / [NH3]²

Kc = (1.5³ * 0.5) / (1²)

Kc = 1.69

The value of K for this reaction is 1.69

Answer:

Scientists have been debating over light being a wave or particle since its recognition.

Sir Issac Newton discovered that light had frequency and other properties. Newton described light to be a particle because it created shadows which were sharp and very clear.

Francesco Maria Grimaldi, claimed that light was a wave. This was because this scientist observed the diffraction of light and hence, claimed light to be a type of wave.

The speed of light is 299 792 458 m / s. Nothing can travel faster than light.

Answer:

Same particles (electrons) were emitted even after changing the cathode material.

Explanation:

In his famous experiment, Thompson tested the properties of atomic particles. He used a cathode ray tube to apply voltage on the cathode. This generated beam of electrons, also called cathode rays. He bombarded the rays on phosphorus on the other end of the tube, to observe the pathway it took.  

When he noticed the deflection of cathode rays when it passes through the electric and magnetic field, he repeated the experiment by changing the cathode material. To his surprise, rays emitted from all the materials exhibited the same behavior.

He concluded that these rays comprising of electrons, are a fundamental part of atoms of every element.  

Answer:

See explanation below

Explanation:

The drawing of the molecule and mechanism, you can see it in the attached pictures.

Now, answering the theorical questions:

The 1-butene is like this:

CH2 = CH - CH2 - CH3

If this molecule reacts with bromine (Br2) the reaction and product formed is:

CH2 = CH - CH2 - CH3 + Br2 -----------> Br-CH2 - CH(Br) - CH2 - CH3

The product formed is called 1,2 - dibromo - butane, and the reaction with halides like bromine is called halogenation. In this case, alkenes halogenation, so, we become a alkene like the 1-butene with a halide like bromine to form an alkane with halides. This reaction is taking place in conditions of Sn1, although this is an addition (Two steps, see picture below for mechanism).

The bromine, has a high electronegativity (2.9) this is even bigger than the iodine (2.7), so, when the bromine acts as a nucleophile in a SN2 or SN1 reaction (like this one),  bromine atom becomes slightly more negative, and iodine atom becomes slightly more positive, so strictly speaking, the molecule is slightly polar. When the difference of the electronegativities is below of 0.4, we can say that the molecule is non-polar.

Because of the explanation above, the reaction is taking place with bromine, because it has a higher electronegativity, even more than the chlorine, so the molecule is more polar and can have a better reaction with the 1-butene than the chlorine. Has a better nucleophyle attack and also, is a great leaving group.

The picture below will show the mechanism:

Answer : The boiling point of the resulting solution is, [tex]100.6^oC[/tex]

Explanation :

Formula used for Elevation in boiling point :

[tex]\Delta T_b=i\times k_b\times m[/tex]

or,

[tex]T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}[/tex]

where,

[tex]T_b[/tex] = boiling point of solution = ?

[tex]T^o_b[/tex] = boiling point of water = [tex]100^oC[/tex]

[tex]k_b[/tex] = boiling point constant  = [tex]0.52^oC/m[/tex]

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

[tex]w_2[/tex] = mass of solute (sucrose) = 5.0 g

[tex]w_1[/tex] = mass of solvent (water) = 10.0 g

[tex]M_2[/tex] = molar mass of solute (sucrose) = 342.3 g/mol

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

[tex](T_b-100)^oC=1\times (0.52^oC/m)\times \frac{(5.0g)\times 1000}{342.3\times (10.0g)}[/tex]

[tex]T_b=100.6^oC[/tex]

Therefore, the boiling point of the resulting solution is, [tex]100.6^oC[/tex]

Answer:

C

Explanation:

Gallium is in group thirteen with outermost electron configuration ns2 np1. The highest occupied sub-level is np1 having only one electron which is the situation required in the question.

Answer:

C.  Ga

Explanation:

Answer:

The ranking is given as; Water > Asphalt > Concrete > glass > Iron

Explanation:

The trick in solving this question is to assume a constant heat value; in this case i'll be choosing 100 J. Use this value to solve for the temperature difference. from that we can be able to rank the samples in order of their temperatures.

The formular to be used here is the;

H = MCΔT

Where;

H = Heat

M = Mass

C = Heat Capacity

ΔT = Temperature difference

ΔT = H/MC

In water;

ΔT = 100 / (10 * 4.184) = 2.39K

In Concrete;

ΔT = 100 / (10 * 0.88) = 11.36K

In asphalt;

ΔT = 100 / (10 * 0.920) = 10.87K

In glass;

ΔT = 100 / (10 * 0.84) = 11.9K

In iron;

ΔT = 100 / (10 * 0.448) = 22.3K

The samples with least temperature difference would have final temperatures and vice versa.

Our ranking is the given as; Water > Asphalt > Concrete > glass > Iron

The ranking from the least final temperature to the greatest is liquid water, asphalt, concrete, glass, iron.

The question involves understanding the concept of specific heat capacity in relation to the final temperature of different materials after the same quantity of heat is transferred. The specific heat capacity (C) is a property that defines how much heat energy is required to raise the temperature of a unit mass of a substance by one degree Celsius. The materials listed are liquid water, concrete, asphalt, glass, and iron, with specific heat capacities of 4.184 J/g°C, 0.88 J/g°C, 0.920 J/g°C, 0.84 J/g°C, and 0.448 J/g°C, respectively.

Given the relationship that the amount of heat (Q) added or removed is directly proportional to the mass (m), specific heat capacity (C), and change in temperature (ΔT), we have Q = mCΔT. With an equal amount of heat transferred and the same mass for each sample, substances with a higher specific heat capacity will experience a smaller change in temperature. Thus, to rank the final temperatures from least to greatest after the equal heat transfer, we should look at the specific heat capacities in reverse order, as a lower specific heat capacity means more temperature change for the same amount of heat.

Iron (C = 0.448 J/g°C), Glass (C = 0.84 J/g°C), Concrete (C = 0.88 J/g°C), Asphalt (C = 0.920 J/g°C), Liquid Water (C = 4.184 J/g°C)

Therefore, the final temperatures of the samples, from least to greatest, will be as follows: iron will have the highest final temperature, followed by glass, concrete, asphalt, and liquid water will have the lowest final temperature.

Answer:

a

Explanation:

Group 15 form trihydrides with the non metal atoms like phosphine, ammonia

Answer:

The product of the reaction between a ketone and an alcohol is initially a hemiketal which yields a ketal on further reaction with another alcohol molecule.

The structure is found in the attachment.

Explanation:

This reaction is a nucleophilic addition to the carbonyl group. In organic chemistry, a nucleophilic addition reaction is an addition reaction where a chemical compound with an electron-deficient or electrophilic double or triple bond, a pi (π) bond, reacts with electron-rich reactant, termed a nucleophile, with the elimination of the double bond and creation of two new single, or sigma (σ), bonds.

In the reaction between a ketone and an alcohol, the carbonyl group of the ketone serves as the electrophile while the hydroxyl group of the alcohol is the nucleophile. The first product is known as a hemiketal because a single alcohol group has been aded to the carbonyl group of the ketone. Further nucleophilic additon of an alcohol group initiated by the presence of an acid e.g hydrochloric acid, results in the formation of a ketal which has two alcohol group added to the original ketone.

Final answer:

The reaction between a ketone and an alcohol can produce a hemiketal or ketal, depending on the reaction conditions and the excess of alcohol. A hemiketal is formed when the alcohol reacts with the ketone to form a new carbon-oxygen bond, while a ketal is formed when a second molecule of alcohol reacts to convert the hemiketal into a stable compound.

Explanation:

In the reaction between a ketone and an alcohol, the product formed is called a hemiketal or ketal, depending on the reaction conditions and the presence of excess alcohol. A hemiketal is formed when the alcohol reacts with the ketone to form a new carbon-oxygen bond, while a ketal is formed when a second molecule of alcohol reacts to convert the hemiketal into a stable compound.

For example, if we take the ketone acetone (CH3C=O) and react it with ethanol (CH3CH2OH), we can form a hemiketal:

CH3C(OC2H5)(OH)

If we add excess ethanol, the hemiketal can react with a second molecule of ethanol to form a ketal:

CH3C(OC2H5)2

The reaction can also occur between other ketones and alcohols, resulting in the formation of different hemiketals or ketals.

Final answer:

In this problem, 1 mole of CO₂ is produced for every mole of carbon atoms and 1 mole of H₂O is produced for every 2 moles of hydrogen atoms. By using these ratios, the masses of carbon and hydrogen in the original sample can be calculated from the masses of CO₂ and H₂O, and their molar masses.

Explanation:

Upon combustion, 1 mol of CO₂ is produced for each mole of carbon atoms in the original sample. Similarly, 1 mol of H₂O is produced for every 2 mol of hydrogen atoms present in the sample. The masses of carbon and hydrogen in the original sample can be calculated from these ratios, the masses of CO₂ and H₂O, and their molar masses. Because the units of molar mass are grams per mole, we must first convert the masses from milligrams to grams:

Answer:

1.90 L

Explanation:

Using Boyle's law  

[tex]{P_1}\times {V_1}={P_2}\times {V_2}[/tex]

Given ,  

V₁ = 595 L  

V₂ = ?

P₁ = 1.00 atm

P₂ = 55.0 atm

Using above equation as:

[tex]{P_1}\times {V_1}={P_2}\times {V_2}[/tex]

[tex]{1.00}\times {595}={55.0}\times {V_2}[/tex]

[tex]{V_2}=\frac{{1.00}\times {595}}{55.0}\ L[/tex]

[tex]{V_2}=1.90\ L[/tex]

The volume would be 1.90 L.

Answer:

1. Dalton..........Father of Atomic theory

2. Bohr..........Postulated the quantum atom

3. nucleus..........location of the most of the mass of the atom

4. Chadwick..........discovered the neutron

5. Rutherford..........discovered the proton

6. electrons..........emitted from a cathode-ray tube

7. J.J. Thomson..........discovered the electron

Explanation:

The question is incomplete.Here is the cmplete question.

Match these items.

1. Dalton...... emitted from a cathode-ray tube

2. Bohr.......... discovered the neutron

3. nucleus.......... discovered the electron

4. Chadwick........ postulated the quantum atom

5. Rutherford........... discovered the proton

6. electrons ............father of atomic theory

7. J. J. Thomson.............location of most of the mass of the atom

1) Dalton is the father of atomic theory

He proposed that matter comprises of indivisible particles called atoms. Atoms are the building block of a matter. All atoms of an element are identical. Atoms of different elements differ from each other in terms of size and mass.

2. Bohr postulated the quantum atom

He proposed that electrons revolve around the nucleus in orbits. Each orbit is labelled by an integer 'n’. This integer is the quantum number. Electrons can move between shells by emitting or absorbing energy.

3. Nucleus is the location of most of the mass of the atom

The entire mass (almost 99%) of atom is concentrated in the nucleus containing protons and neutrons. Electrons orbiting around have negligible mass compared the protons and neutrons.  

4. Chadwick discovered the neutron.

In an experiment, Chadwick bombarded beryllium atoms with alpha rays. He noticed that beryllium emitted neutral rays as a result. Unlike gamma rays, the rays did not create photo electric effect when they hit charged electroscope. He concluded that they are neutrons instead.  

5. Rutherford discovered the proton  

In his famous gold foil experiment, he bombarded positively charged alpha rays to gold foil as saw a large proportion of them being deflected. He concluded that the atoms must have positively charged particles that caused the deflection.

6. Electrons are emitted from a cathode-ray tube  

When electricity is passed through the cathode in the tube, electrons in the outermost orbit gain enough energy to break out from it

7. J. J. Thomson discovered the elections.  

In the cathode-ray tube, Thompson observed that the rays emitted from the cathode are deflected towards to the positively charged plate. He concluded that cathode rays composed of negatively charged particles, i.e. electrons.  

Answer:

Correct matches below.

Explanation:

Dalton - Father of Atomic Theory

Chadwick - Discovered the neutron

J.J Thomson - Discovered the electron

Bohr - Postulated the quantum atom

Rutherford - Discovered the proton

Nucleus - Location of most of the mass in the atom

Electrons - Emitted from a cathode-ray tube

Answer:

The percent by mass of nitric acid in the mixture is 11.1 %

Explanation:

Step 1: Data given

Mass of HNO3 = 8.89 grams

Volume of KOH = 27.1 mL = 0. 0271 L

Molarity of KOH = 0.581 M

Step 2: The balanced equation

HNO 3  +  KOH  →  KNO 3  +  H 2 O

Step 3: Calculate the moles of KOH

Moles of KOH = molarity KOH * volume

Moles KOH = 0.581 M * 0.0271 L

Moles KOH = 0.0157 moles

Step 4: Calculate moles of HNO3

For 1 mol of KOH we need 1 mol of HNO3

For 0.0157 moles of KOH we need 0.0157 moles of HNO3

Step 5: Calculate mass of HNO3

Mass KOH = moles KOH * molar mass KOH

Mass KOH = 0.0157 moles * 63.01 g/mol

Mass KOH = 0.989 grams

Step 6: Calculate mass % HNO3 in sample

mass % = (0.989 grams / 8.89 grams)*100%

mass % = 11.1 %

The percent by mass of nitric acid in the mixture is 11.1 %

The molarity of the 360mg aspirin sample dissolved in 200mL solution is found to be 0.01 M. As molarity is a measure of concentration, it remains the same in a 50mL sample of the solution. Therefore, the molarity of the aspirin in the 50mL solution is also 0.01 M.

To calculate the molarity of the aspirin in a 50mL sample, first the molarity of the original 200mL solution is calculated. The molarity (M) is defined as moles of solute (in this case aspirin) per liters of solution. The moles of aspirin in the 360mg sample can be calculated by dividing by the molar mass of aspirin, which is 180g/mol. Thus, there are 0.002 mol (360mg * 1g/1000mg * 1 mol/180g) of aspirin in the 200mL solution. Converting mL to L (200mL * 1L/1000mL), the molarity of the 200mL solution is 0.002 mol / 0.2 L = 0.01 M.

Since molarity is a concentration, it remains the same regardless of the volume of the solution: thus, the molarity of the 50mL sample of the solution is also 0.01 M. So, the molarity of aspirin in a 50mL sample of the solution is 0.01 M.

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Answer: If all of the carbon atoms are linked by single covalent bonds and there are no branches, the compounds are called homologous series.

Explanation:

A series of carbon atoms which include different number of carbon atoms but have same functional group are known as homologous series.

Generally, these type of series have a chemical formula as [tex]C_{n}H_{2n+2}[/tex].

No branches are present in this type of series.

For example, [tex]CH_{4}[/tex], [tex]C_{2}H_{6}[/tex], [tex]C_{3}H_{8}[/tex] etc are all homologous series.

Thus, we can conclude that if all of the carbon atoms are linked by single covalent bonds and there are no branches, the compounds are called homologous series.

There are 1.25 moles of oxygen atoms in 20 g of O2, calculated by converting the mass to moles using the molar mass.

To determine the number of moles of oxygen atoms in 20 g of [tex]\(O_2\)[/tex], we first need to find the molar mass of [tex]\(O_2\)[/tex]. Oxygen [tex](\(O\))[/tex] has an atomic mass of approximately 16 g/mol. Since [tex]\(O_2\)[/tex] molecules contain two oxygen atoms, the molar mass of [tex]\(O_2\) is \(2 \times 16 \, \text{g/mol} = 32 \, \text{g/mol}\).[/tex]

Next, we use the formula:

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

Substituting the given mass of [tex]\(20 \, \text{g}\)[/tex] and the molar mass of [tex]\(O_2\) (\(32 \, \text{g/mol}\)):[/tex]

[tex]\[ \text{Number of moles} = \frac{20 \, \text{g}}{32 \, \text{g/mol}} \][/tex]

[tex]\[ \text{Number of moles} = 0.625 \, \text{mol} \][/tex]

Since each molecule of [tex]\(O_2\)[/tex] contains 2 oxygen atoms, the number of moles of oxygen atoms is twice the number of moles of [tex]\(O_2\)[/tex]:

[tex]\[ \text{Number of moles of oxygen atoms} = 2 \times 0.625 \, \text{mol} = 1.25 \, \text{mol} \][/tex]

Therefore, there are [tex]\(1.25 \, \text{mol}\)[/tex] of oxygen atoms in [tex]\(20 \, \text{g}\)[/tex] of [tex]\(O_2\)[/tex].

The question probable maybe:

How many moles of oxygen atoms are there in 20 g of O2?

Answer:

The coefficient of oxygen in the balanced equation is equal to 3.

Explanation:

The combustion reaction of ethanol in the "gasohol" produces carbon dioxide and water as follows:        

CH₃CH₂OH + O₂ → CO₂ + H₂O       (1)      

To find the coefficient of oxygen in equation (1), we need to balance it. The balanced reaction is the next:

CH₃CH₂OH + 3O₂ → 2CO₂ + 3H₂O      

In the balanced equation, we have the same number of carbon, hydrogen, and oxygen atoms in the products than in the reactants. Therefore, the coefficient of oxygen in the balanced equation is equal to 3.  

I hope it helps you!          

Final answer:

The coefficient of oxygen in the balanced chemical equation for the combustion of ethanol is 3.

Explanation:

The combustion of ethanol, C₂H₅OH, when it is blended with gasoline to make 'gasohol' involves a reaction with oxygen to produce carbon dioxide and water.

The balanced chemical equation for this combustion is C₂H₅OH(l) + 3O₂(g) --> 2CO₂(g) + 3H₂O(g).

Therefore, the coefficient of oxygen in the balanced equation is 3, indicating that three molecules of diatomic oxygen are required for the complete combustion of one molecule of ethanol.

Answer:

The answer to your question is letter A

Explanation:

Process

1.- Calculate the molar mass of KNO₃

KNO₃    molecular mass = 39.1 + 14.01 + (3 x 16)

                                        = 39.1 + 14.01 + 48

                                        = 101.11 g

2.- Use a rule of three to find the percent of nitrogen

                            101.11 g of KNO₃  ---------------   100%

                             14.01 g of N        ---------------     x

                             x = (14.01 x 100) / 101.11

                             x = 13.86%

Answer: D

Explanation:

The molecular theory of gases states that there are no intermolecular forces between gases. Gas molecules are separated from each other such that individual molecules are far apart from each other. When volume is increased, gas molecules spread out from each other and the distance between them increases thus approximating the situation in ideal gases.

Final answer:

The behavior of a sample of a real gas more closely approximates that of an ideal gas as its volume is increased because the average distance between molecules becomes greater.

Explanation:

The behavior of a sample of a real gas more closely approximates that of an ideal gas as its volume is increased at constant temperature because the average distance between molecules becomes greater. In an ideal gas, the molecules are assumed to have zero volume, while in real gases, the molecules have small but measurable volumes. As the volume of the gas increases, the intermolecular distances become larger, reducing the frequency of molecule-wall collisions. This behavior is described by Avogadro's law, which states that increasing the number of gas molecules requires a proportional increase in the container volume to yield a constant number of collisions per unit wall area per unit time.

Answer and Explanation:

In order to predict the feasibility of redox processes, standard electrode potentials are majorly employed. Generally, if the electrode potential for the reaction is positive, it is considered to be feasible. However, some conditions affect this statement

The value of E° talks about the feasibility of the reaction under standard conditions only and says nothing about the reaction rate.

A positive value of E° means, the equilibrium constant K is greater than 1; while a negative value of E° means, that it is less than 1.

The attachment below shows the simple analysis of the feasibility of two different reactions A and B, at standard and non standard conditions respectively.

NOTE: Standard conditions for Redox reaction: 298.15K(Temperature), 1 atm(Pressure), 1.0M(Concentration) for both anode and cathode.

Non standard conditions for Redox reaction: Any of the 3 conditions above are changed, especially the concentration.

Answer:

type 2 diabetes

insulin

Explanation:

type 2 diabetes is a chronic condition that affects the way the body processes blood sugar. A patient with type 2 diabetes in the body either doesn't produce enough insulin, or it resists insulin.

As Chromium levels can be below normal in people with type 2  diabetes. Research studies shows that taking drugs that contains chromium such as chromium picolinate can help increase the effectiveness of insulin levels and help insulin work in people with type 2 diabetes.

Answer:

[tex]\rho=1.54\ g/cm^3[/tex]

Explanation:

The expression for density is:

[tex]\rho=\frac {Z\times M}{N_a\times {{(Edge\ length)}^3}}[/tex]

[tex]N_a=6.023\times 10^{23}\ {mol}^{-1}[/tex]

M is molar mass of Calcium = 40.078 g/mol

For cubic closest packed structure , Z= 4

[tex]\rho[/tex] is the density

Radius = 197 pm = [tex]1.97\times 10^{-8}\ cm[/tex]

Also, for fcc, [tex]Edge\ length=2\sqrt{2}\times radius=2\sqrt{2}\times 1.97\times 10^{-8}\ cm=5.572\times 10^{-8}\ cm[/tex]

Thus,  

[tex]\rho=\frac{4\times \:40.078}{6.023\times \:10^{23}\times \left(5.572\times 10^{-8}\right)^3}\ g/cm^3[/tex]

[tex]\rho=\frac{160.312}{10^{23}\times \:6.023\left(10^{-8}\times \:5.572\right)^3}\ g/cm^3[/tex]

[tex]\rho=\frac{160.312}{10^{23}\times \:1.04195E-21}\ g/cm^3[/tex]

[tex]\rho=\frac{160.312}{104.19483}\ g/cm^3[/tex]

[tex]\rho=1.54\ g/cm^3[/tex]

The density of solid calcium can be calculated by determining the density of its unit cell using the face-centered cubic (FCC) structure. The mass and volume of the unit cell can be calculated using the atomic radius and atomic mass of calcium. Dividing the mass by the volume gives the density of solid calcium.

The density of solid calcium can be calculated by determining the density of its unit cell, which is a face-centered cubic (FCC) structure. In an FCC structure, each unit cell contains 4 atoms. The mass of 4 calcium atoms can be calculated using the atomic mass of calcium, and the volume of the unit cell can be calculated using the atomic radius of calcium. Dividing the mass by the volume gives the density of solid calcium.

The atomic radius of calcium is given as 197 pm, which can be converted to cm by multiplying by 10^-10. The volume of the unit cell can be calculated using the formula V = (edge length)^3. The edge length can be calculated using the diagonal of the face, which is 4 times the atomic radius. The mass of 4 calcium atoms can be calculated using the atomic mass of calcium, which is 40.08 g/mol. Dividing the mass by the volume gives the density of solid calcium.

Density of solid calcium = mass of 4 Ca atoms / volume of unit cell

Keywords: density, solid calcium, unit cell, face-centered cubic (FCC) structure, atomic radius, atomic mass

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Answer:

Capable of bonding to surfaces with the application of light pressure is a chemical reaction.

Explanation:

Here, bonding occurs . Bonding results in destruction of old bond and formation of new bonds.Hence new substance with completely different properties is formed. These changes occur only in a chemical reaction .

In physical processes no new substance is formed (no bonding).So,no change in properties of a substance . ]

This process (capability of bonding to surfaces with the application of light pressure) results in chemical reaction.

This type of substances are called Pressure-sensitive Adhesives.(PSA)

Answer:

Moles of potassium chlorate = 0.02976 moles

Explanation:

At standard pressure and temperature,

22.4 L of a gas consists of 1 mole

Thus, given, volume of [tex]O_2[/tex] = 1.0 L

So,

1 L of a gas consists of [tex]\frac{1}{22.4}[/tex] mole

Moles of oxygen gas = 0.04464 moles

The reaction is shown below as:-

[tex]2KClO_3\rightarrow 2KCl+3O_2[/tex]

3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.

So,

1 mole of oxygen gas are produced when [tex]\frac{2}{3}[/tex] moles of potassium chlorate undergoes reaction.

Thus,

0.04464 mole of oxygen gas are produced when [tex]\frac{2}{3}\times 0.04464[/tex] moles of potassium chlorate undergoes reaction.

Moles of potassium chlorate = 0.02976 moles

From the decomposition reaction 2KClO₃(s) → 2KCl(s) + 3O₂(g), the number of moles of KClO₃ to be decomposed to generate 1.0 L of O₂ gas at standard temperature and pressure (STP) is 0.030.

The balanced chemical reaction for the decomposition of potassium chlorate (KClO₃) is the following:

2KClO₃(s) → 2KCl(s) + 3O₂(g)   (1)

We can find the number of moles of O₂ gas with the Ideal gas equation:

[tex] PV = nRT [/tex]

Where:

P: is the pressure = 1.0 atm (at STP conditions)

V: is the volume = 1.0 L

R: is the gas constant = 0.082 L*atm/(K*mol)

T: is the temperature = 273 K (at STP conditions)

n: is the number of moles =?

The number of moles of O₂ gas is:

[tex] n_{O_{2}} = \frac{PV}{RT} = \frac{1.0 atm*1.0 L}{0.082 L*atm/(K*mol)*273 K} = 0.045 \:moles [/tex]

From reaction (1), we have that 2 moles of KClO₃ produce 3 moles of O₂, so the number of moles of KClO₃ resulting from the decomposition is:

[tex] n_{KClO_{3}} = \frac{2\:moles\:KClO_{3}}{3\:moles\:O_{2}}*0.045\:moles\:O_{2} = 0.030 \:moles [/tex]

Therefore, the number of KClO₃ moles to be decomposed is 0.030.

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