Calculate the entropy change for the reaction: HCl(g) + NH3(g) -> NH4Cl(s) Entropy data: HCl: 187 J/K mol NH3: 193 J/K mol NH4Cl: 94.6 J/K mol

Answer:

1) before the addition of barium hydroxide

pH = -log[H⁺] = -log (0.276) = 0.559≈0.56

2)after the addition of barium hydroxide

pH = -log [H⁺] = -log(0.0857) = 1.067

3)at equivalent point, the solution will be neutral

pH = 7.0

4) after adding 40.7mL barium hydroxide

Explanation:

equation of reaction

2HCl(aq) + Ba(OH)₂(aq) ------->BaCl₂(aq) + 2H₂O(l)

1) Before the addition of barium hydroxide

concentration of HBr = 0.276M

[H⁺] = 0.276M

pH = -log[H⁺] = -log (0.276) = 0.559≈0.56

2) After adding 16.7mL barium hydroxide

moles of [OH⁻] = 16.7mL × 0.105 × 2

=3.507m mol = 3.507 × 10³mol

moles  of [H⁺] = 25.5mL × 0.276M

=7.038m mol = 7.038 × 10³mol

moles of  [H⁺] remaining = (7.038 - 3.421)m mol

= 3.617m mol = 3.617 × 10³mol

[H⁺]= [tex]\frac{3.617}{25.5 + 16.7}[/tex] = 0.0857

pH = -log [H⁺] = -log(0.0857) = 1.067

3) At equivalent point, the solution will be neutral

pH = 7.0

4) After adding 40.7mL barium hydroxide

moles of [OH⁻] = 40.7mL × 0.105M × 2

=8.547

moles of [OH⁻] remaining = 8.547 - 7.038

= 1.509m mol = 1.509 × 10³mol

pOH= -log[OH⁻]= 2.82

pH = 14 - 2.82 = 11.18

The question is incomplete, the complete question is;

What is the purpose of reacting 2.0mL of HNO3 with 2.0 mL of H2SO4 in a separate test tube, prior to adding it to the solution containing the substrate? more than one answer is possible

A) The release of a water molecule that acts as an electrophile in the reaction with methyl benzoate.

B) The formation of nitronium ion, which acts an electrophile in the reaction with methylbenzoate.

C)The formation of bisulfate (hydrogen sulfate), which acts as an electrophile in the reaction with methylbenzoate.

D)The release of a water molecule that acts as a nucleophile in the reaction with methyl benzoate.

Answer:

B) The formation of nitronium ion, which acts an electrophile in the reaction with methylbenzoate.

Explanation:

The benzene ring is known to be stable hence it can only undergo a substitution reaction with the aromatic ring still intact. When the substitution reaction involves an electrophile we refer to the process as electrophillic aromatic substitution. Electrophilic aromatic substitution is a useful synthetic route for many organic compounds.

In the electrophilic substitution of methyl benzoate using the 1:1 volume ratio mixture of H2SO4/HNO3, the nitronium ion (NO2+) is the electrophile generated in the test tube. It is this NO2+ that now reacts with the methyl benzoate to yield the reaction product.

Answer:

C. To determine how efficient reactions are.

D. To determine how much reactant they need.

Explanation:

When you are doing a reaction, you are hoping for a percent yield to close of 100%. You make the reaction and determine how many product you obtain. If you know the percent yield of a reaction you can calculate the amount of reactant you need to obtain a determined amount of product.

Having this in mind:

A. To balance the reaction equation.  false. To calculate percent yield you need to balance the reaction before. You don't use percent yield to balance the reaction

B. To determine how much product they will need.  false. You determine how much product you obtain after the reaction. How much product you need is independent of percent yield

C. To determine how efficient reactions are.  true. A way to determine efficience of a reaction is with percent yield. An efficient reaction has a high percent yield.

D. To determine how much reactant they need. true. If you know percent yield of a reaction you can know how many reactant you must add to obtain  the amount of product you want.

Answer:

Benzoic acid: 1.288g

Sodium benzoate: 124.48g

Explanation:

Benzoic acid, HC7H5O2 is in equilibrium with its conjugate base, C7H5O2⁻ producing a buffer. The pH of the buffer can be determined following H-H equation:

pH = pKa + log [C7H5O2⁻] / [HC7H5O2] (1)

Where pH is desire pH = 6.10 pKa is -log Ka = 4.187 and [] are molar concentrations of the buffer.

As you want to prepare 2.5L of a 0.35M of buffer, moles of buffer are:

2.5L ₓ (0.35mol / L) = 0.875moles of buffer.

And you can write:

0.875 moles = [C7H5O2⁻] + [HC7H5O2] (2)

Replacing (2) in (1)

pH = pKa + log [C7H5O2⁻] / [HC7H5O2]

6.10 = 4.187 + log [C7H5O2⁻] / [HC7H5O2]

1.913 =  log [C7H5O2⁻] / [HC7H5O2]

81.846 = 0.875mol - [HC7H5O2] / [HC7H5O2]

81.846 [HC7H5O2] = 0.875mol - [HC7H5O2]

82.846 [HC7H5O2] = 0.875mol

And moles of the benzoate, [C7H5O2⁻]:

[C7H5O2⁻] = 0.875mol - 0.01056mol =

Using molar mass of benzoic acid and sodium benzoate, amount of each compound you must add to prepare 2.5L of the buffer are:

Benzoic acid: 0.01056mol ₓ (122.01g/mol) = 1.288g

Sodium benzoate: 0.8644mol ₓ (144.01g/mol) = 124.482g

Answer: The percentage composition of nitrogen , hydrogen and oxygen is 22.2 % , 1.59 % and 76.2% respectively.

Explanation:

Percentage composition is defined as the ratio of mass of substance to the total mass in terms of percentage.

Percentage composition=[tex]\frac{\text {mass of the element}}{\text {Total mass of the substance}}\times 100\%[/tex]

a) [tex]{\text {percentage composition of nitrogen}}=\frac{\text {mass of nitrogen}}{\text {Total mass}}\times 100\%[/tex]

[tex]{\text {percentage composition of nitrogen}}=\frac{9.8g}{9.8+0.7+33.6}\times 100\%=22.2\%[/tex]

b) [tex]{\text {percentage composition of hydrogen}}=\frac{\text {mass of hydrogen}}{\text {Total mass}}\times 100\%[/tex]

[tex]{\text {percentage composition of hydrogen}}=\frac{0.7}{9.8+0.7+33.6}\times 100\%=1.59\%[/tex]

c) [tex]{\text {percentage composition of oxygen}}=\frac{\text {mass of oxygen}}{\text {Total mass}}\times 100\%[/tex]

[tex]{\text {percentage composition of oxygen}}=\frac{33.6}{9.8+0.7+33.6}\times 100\%=76.2\%[/tex]

The percentage composition of nitrogen , hydrogen and oxygen is 22.2 % , 1.59 % and 76.2% respectively.

Answer:

They are called nodes.

Explanation:

Answer:

i guess this is the ans nodes

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Answer:

a. increase

Explanation:

Catalysis is the process of increasing the rate of a chemical reaction by adding a substance known as a catalyst, which is not consumed in the catalyzed reaction.

By default, catalysts exists to speed up the rate of reactions. Increasing the amount of catalysts means that there would be an increase in the rate of reaction. The correct option is A.

Answer:

V2 = 17371.43ml

Explanation:

We use Boyles laws

since temperature is constant

P1V1=P2V2

760 x 400 = 17.5 x V2

304000 = 17.5 x V2

V2 = 304000/17.5

V2 = 17371.43ml

The volume will the dry hydrogen gas occupy at the temperature of 20°C and vapor pressure at  760 torrs will be 18 ml.

What is vapor pressure?

The vapor pressure of a liquid is independent of the volume of liquid in the container, whether one liter or thirty liters; both samples will have the same vapor pressure at the same temperature.

The temperature has an exponential connection with vapor pressure, which means that as the temperature rises, the vapor pressure rises as well the equation is -

P1 V1 / T1 = P2 V2 / T1

here, P = pressure

       T = temperature

        V = volume

substituting the value in the equation,

400 ×760 / 20 = 17.5× V / 20

V = 400× 760 / 20 × 17.5 / 20

V = 18 ml

Therefore the volume of the hydrogen gas remaining at this temperature will be 18 ml.

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Answer:

Chemical reaction B governs the process

Explanation:

The first part of the question asks to convert the mass of the calcium carbonate given to number of moles.

Mathematically;

Number of moles = mass/molar mass

Molar mass of CaCO3 = 100 g/mol

So the number of moles of CaCO3 will be 2.49/100 = 0.0249 moles

The second part of the question asks to convert the mass of carbon iv oxide to moles of carbon iv oxide

Mathematically;

That is same as ;

Number of moles = mass/molar mass

molar mass of CO2 is 44 g/mol

Number of moles of CO2 = 1.13/44 = 0.0256 moles

Now, if we compare the values of these number of moles, we can see that there are almost equal.

What this means is that the number of moles of calcium carbonate reacted is equal to the number of moles of carbon iv oxide produced.

So what we conclude here is that we have an equal mole ratio between the two compounds.

So the reaction that would be the correct answer will present equal number of moles of carbon iv oxide and calcium carbonate

Thus, we can see that reaction B is the one that governs this process as it is the only reaction out of the three options that present the two compounds with equal number of moles.

Answer:

f is =2,2-dimethyl butane

g is = 2,2-dimethyl propane

h is = 3,3-diethyl pentane

Explanation:

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✔ C5H4 has a molecular molar mass of :

M(C5H4) = 5 x M(C) + 4 x M(H)

✔ The molecular mass of C5H4 is therefore 64 g/mol.

Answer:

C10H8

Explanation:

I clicked on that answer, and it is correct.

Answer:

Answers are in the explanation.

Explanation:

Given concentrations are:

In the reaction:

2SO₂(g) + O₂(g) ⇄ 2SO₃(g)

Kc is defined as:

Kc = 15 = [SO₃]² / [O₂] [SO₂]²

Where concentrations of each species are equilbrium concentrations.

Also, you can define Q (Reaction quotient) as:

Q = [SO₃]² / [O₂] [SO₂]²

Where concentrations of each species are ACTUAL concentrations.

If Q > Kc, the reaction will shift to the left until Q = Kc;

If Q < Kc, the reaction will shift to the right until Q = Kc

If Q = Kc, there is no net reaction because reaction would be en equilibrium.

Replacing with given concentrations:

Answer:

3.00 L

Explanation:

PV = nRT

(82 atm × 101325 Pa/atm) V = (10 mol) (8.314 J/mol/K) (27 + 273) K

V = 0.00300 m³

V = 3.00 L

Answer:

Use 62% - the equation is for the amount present at a given time. 0.62 = (1) e-kt -> ln(0.62)=-kt -> k = -ln(0.62)/t. I get k = .00117 hr-1 t(half) = 0.693/k = 590 hr.

HOPE THIS HELPS AND PLSSS MARK AS BRAINLIEST AND THNXX :)

The half-life is the time at which the substance's concentration is reduced by half of its initial amount. The half-life of a nuclide that lost its 38.0% mass is 590 hr. Thus, option C is correct.

Half-life is the time required by a substance to get reduced to half of its initial concentration. The half-life of the substance can be determined by the rate constant.

Given,

The initial quantity of substance (A₀) = 100

Remaining quantity (At) = 10 - 38 = 62

Time elapse (t) = 407 hours

The rate constant (k) is calculated as:

ln (At ÷ A₀) = - kt

ln (62 ÷ 100) ÷ 407 hour = - k

-0.47803580094 ÷ 407 = - k

k = 0.00117453513

Now, half-life from rate constant (k) is calculated as:

[tex]\rm t ^{\frac{1}{2}}[/tex] = 0.693 ÷ k

[tex]\rm t ^{\frac{1}{2}}[/tex] = 0.693 ÷ 0.00117453513

[tex]\rm t ^{\frac{1}{2}}[/tex] = 590 hours

Therefore, option C. 590 hours is the half-life of the substance.

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Explanation:

Hi there!

I attached a photo of a unit summary that states the difference between s-t and v-t graph.

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Answer:

Explanation:

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

ionisation constant = 1.36 x 10⁻⁴ .

molecular weight of lactic acid = 90 g

moles of acid used = 20 / 90

= .2222

it is dissolved in one litre so molar concentration of lactic acid formed

C = .2222M

Let n be the fraction of moles ionised  

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

C  - nC                                          nC                  nC

By definition of ionisation constant Ka

Ka = nC x nC / C - nC

= n²C ( neglecting n in the denominator )

n² x .2222 = 1.36 x 10⁻⁴

n = 2.47  x 10⁻²

nC = 2.47  x 10⁻² x .2222

= 5.5 x 10⁻³

So concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per litre .

The concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per liter .

Ionization of lactic acid can be represented as:

CH₃CHOHCOOH⇄ CH₃CHOHCOO⁻  + H⁺

Given:

ionization constant = 1.36 x 10⁻⁴

mass= 20.0 g

Now, Molecular weight of lactic acid = 90 g

[tex]\text{Number of moles}=\frac{20}{90} =0.22mol[/tex]

It is dissolved in 1.00L so molar concentration of lactic acid formed will be

C = 0.22M

Consider "n" to be the fraction of moles ionized  

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

C  - nC                                          nC                  nC

By definition of ionization constant Ka

[tex]K_a =\frac{nC*nC}{C-nC}[/tex]

[tex]K_a= n^2C[/tex] ( neglecting n in the denominator )

On substituting the values we will get:

[tex]n^2 *0.22 = 1.36 *10^{-4}\\\\n = 2.47 * 10^{-2}[/tex]

To find the concentration of hydronium ion in the solution,

[tex]nC = 2.47 *10^{-2} *0.22\\\\nC= 5.5 * 10^{-3}[/tex]

So, concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per liter.  

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Answer:

1. 6.92 g of H2O

2i. - 46 KJ of energy.

ii. Option A. Evolved.

Explanation:

1. Determination of the mass of H2O that would be made to react if 110 kJ of energy were provided.

This can be obtained as follow:

The equation for the reaction is given below

2H2O(l) —> 2H2(g) + O2(g) H = 572 kJ

Next, we shall determine the mass of H2O required to produce 572 kJ from the balanced equation.

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36 g

From the balanced equation above, 36 g of H2O reacted to produce 572 kJ of energy.

Finally, we shall determine the mass of water (H2O) needed to produce 110 kJ of energy.

This is illustrated below:

From the balanced equation above, 36 g of H2O reacted to produce 572 kJ of energy.

Therefore, Xg of H2O will react to 110 kJ of energy i.e

Xg of H2O = (36 x 110)/572

Xg of H2O = 6.92 g

Therefore, 6.92 g of H2O is needed to react in order to produce 110 KJ of energy.

2i. Determination of the energy.

The balanced equation for the reaction is given below:

CO(g) + 3H2(g) —> CH4(g) + H2O(g) H = -206 kJ

Next, we shall determine the mass of CO that reacted to produce -206 kJ of energy from the balanced equation.

This is illustrated below:

Molar mass of CO = 12 + 16 = 28 g/mol

Mass of CO from the balanced equation = 1 x 28 = 28 g

From the balanced equation above,

28 g of CO reacted to produce -206 kJ of energy.

Finally, we shall determine the amount of energy produced by reacting 6.27 g of CO. This is illustrated below:

From the balanced equation above,

28 g of CO reacted to produce -206 kJ of energy.

Therefore, 6.27 g of CO will react to produce = (6.27 x -206)/28 = - 46 KJ of energy.

Therefore, - 46 KJ of energy were produced from the reaction.

2ii. Since the energy obtained is negative, it means heat has been given off to the surroundings.

Therefore, the heat is evolved.

Answer:

It usually consists of six steps: question, observation or investigation, hypothesis, experiment, analysis of data (reviewing what happened during the experiment), and conclusion. Scientific inquiry, on the other hand, is non-linear, which means it does not follow a consistent step-by-step process.

Explanation:

Hope it helps

Answer:

4NH₃(g)  + 5O₂(g)  →  4NO(g)  +  6H₂O

2NO(g) + O₂(g) → 2 NO₂

Explanation:

First of all, we need to consider the reaction for production of ammonia. In this reaction we have as reactants, nitrogen and hydroge.

3H₂ (g) +  N₂(g)  →  2NH₃ (g)

Afterwards, ammonia reacts to oxygen, to produce NO and H₂O

The equation for the process will be:

4NH₃(g)  + 5O₂(g)  →  4NO(g)  +  6H₂O

Then, we take the nitric oxide to make it react, to produce NO₂, in order to produce nitric acid, for the final reaction:

2NO(g) + O₂(g) → 2 NO₂

3NO₂(g) + H₂O(g) → 2 HNO₃ (g) + NO(g)

Answer:

Molar mass of the gas is 0.0961 g/mol

Explanation:

The effusion rate of an unknown gas = 11.1 min

rate of [tex]H_{2}[/tex] effusion = 2.42 min

molar mass of hydrogen = 1 x 2 = 2 g/m

molar mas of unknown gas = ?

From Graham's law of diffusion and effusion, the rate of effusion and diffusion is inversely proportional to the square root of its molar mass.

from

[tex]\frac{R_{g} }{R_{h} }[/tex] = [tex]\sqrt{\frac{M_{h} }{M_{g} } }[/tex]

where

[tex]R_{h}[/tex] = rate of effusion of hydrogen gas

[tex]R_{g}[/tex] = rate of effusion of unknown gas

[tex]M_{h}[/tex] = molar mass of H2 gas

[tex]M_{g}[/tex] = molar mass of unknown gas

substituting values, we have

[tex]\frac{11.1 }{2.42 }[/tex] = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]

4.587 = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]

[tex]\sqrt{M_{g} }[/tex] = [tex]\sqrt{2}[/tex]/4.587

[tex]\sqrt{M_{g} }[/tex] = 0.31

[tex]M_{g}[/tex] = [tex]0.31^{2}[/tex] = 0.0961 g/mol

The molar mass of the unknown gas will be "0.0961 g/mol".

Given:

Effusion rate of unknown gas,

Effusion rate of [tex]H_2[/tex],

Molar mass of hydrogen,

              [tex]= 2 \ g/m[/tex]

According to the Graham's law, we get

→    [tex]\frac{R_g}{R_h} = \sqrt{\frac{M_h}{M_g} }[/tex]

By substituting the values, we get

→   [tex]\frac{11.1}{2.42} = \sqrt{\frac{2}{M_g} }[/tex]

→ [tex]4.587=\sqrt{\frac{2}{M_g} }[/tex]

→ [tex]\sqrt{M_g} = \sqrt{\frac{2}{4.587} }[/tex]

   [tex]\sqrt{M_g} = 0.31[/tex]

       [tex]M_g = 0.0961 \ g/mol[/tex]

Thus the above solution is right.          

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Answer:

The concentration of one or more of the products is small.

The reaction will not proceed very far to the right.

The reaction will generally form more reactants than products  

Explanation:

We often write

K =[Products]/[Reactants]

Thus, if K is small

A. is wrong. Usually, if K < 1, the concentration of reactants is greater than that of the products.

Answer:

Lewis acid

Explanation:

In chemistry, a Lewis acid is any chemical specie that accepts a lone pair of electrons while a Lewis base is any chemical specie that donates a lone pair of electrons.

If we look at the formation of PF6^-, the process is as follows;

PF5 + F^- -----> PF6^-

We can see that PF5 accepted a lone pair of electrons from F^- making PF5 a lewis acid according to our definition above.

Hence in the formation of PF6^-, PF5 acts a Lewis acid.

Answer:

16 J/K.mol

Explanation:

From the question,

ΔS = ΔH/T............... Equation 1

Where ΔH = Heat change, T = Temperature

But,

ΔH = n(Hv).................. Equation 2

Where n = number of  mole, Hv = heat of vaporization.

Given: Hv = 28.0 kJ/mol, n = 473/85 = 5.59 mole.

Substitute these values into equation 2

ΔH = 28/5.59

ΔH = 5.01 kJ.

Also: T = 273+39.8 = 312.8 J

Substitute into equation 1

ΔS = 5.01/312.8

ΔS = 0.016 kJ/K

ΔS = 16 J/K.mol

Answer

Hello

I think the reaction is like this FeO+OFe²O³

And the balance reaction is 2Fe+OFe²O³

Explanation:

At first we should find sth that has more atoms than the other then for example we realized that we have two atoms of Fe in Fe²O³ then put 2 before FeO and now we have 2 atoms of Fe in right side and 2 atoms of Fe in left then Oxygen in FeO change to 2 atoms of Oxygen and we have an other one in right side that they become 3 atoms of Oxygen and now we have 3 atoms of Oxygen in both right and left side.

Finally our reaction balanced.

Good luck

Explanation:

Answer:

[tex]\boxed{\sf Option \ E}[/tex]

Explanation:

All the gases at the same temperature and mass have the same average kinetic energy.

If the masses were different, then the different gases will have different velocities. If the temperature was higher then there would be a greater motion, if the temperature was lower, then there would be less motion.

Answer:

option E

Explanation:

Answer:

8.96g \ cm3

Explanation:

(89.6\ 10) (g\ cm3) = 8.96g\cm3

Answer:

592 K or 319° C

Explanation:

From the statement of Charles law we know that the volume of a given mass of gas is directly proportional to its absolute temperature at constant pressure. Thus;

V1/T1= V2/T2

Initial volume V1 = 1.75 L

Initial temperature T1= 23.0 +273 = 296 K

Final volume V2= 3.50 L

Final temperature T2 = the unknown

T2= V2T1/V1= 3.50 × 296 / 1.75

T2 = 592 K or 319° C

Answer:

D

Explanation:

Answer:

THE MOLAR MASS OF THE GAS IS 147.78 G/MOLE

Explanation:

Using PV = nRT

n = Mass / molar mass

P = 732.6 mmHg = 1 atm = 760 mmHg

So therefore 732.6 mmHg will be equal to 732.6 / 760 = 0.964 atm

P = 0.964 atm

V = 275 mL = 275 *10 ^-3 L

R = 0.082 Latm/ mol K

T = -28 C = 273 - 28 K = 245 K

mass =  1.95 g

molar mass = unknown

Having known the other variables in the formula, the molar mass of the gas can be obtained.

PV = m R T/ molar mass

Molar mass = m RT / PV

Molar mass = 1.95 * 0.082 * 245 / 0.964 * 275 *10^-3

Molar mass = 39.1755 / 265.1 *10^-3

Molar mass = 39.1755 / 0.2651

Molar mass = 147.78 g/mol

The molar mass of the gas is 147.78 g/mol