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elena-14-01-66 [18.8K]
3 years ago
8

a factor of 8, it means that A is now only 1/8 of its original concentration. A first order reaction, where A → products, has a

rate constant of 1.56 × 107 s −1 . At some time, a concentration of 1.06 × 10−6 M of species A is introduced into the reactor. How long does it take for the concentration of A to fall by a factor of 8?
Chemistry
1 answer:
Trava [24]3 years ago
4 0

Answer:

It takes 1.32x10⁻⁷s for the concentration of A to fall by a factor of 8

Explanation:

The equation that represents a first-order kinetics is:

Ln ([A] / [A]₀] = -kt

<em>Where [A] is actual concentration, [A]₀ is initial concentration, K is rate constant (For the given problem, 1.57x10⁷s⁻¹ and t is time.</em>

<em />

As you want the time when you have [A] in a factor of 8 = [A] / [A]₀ = 1/8

Replacing:

Ln ([A] / [A]₀] = -kt

Ln (1/8) = -1.57x10⁷s⁻¹*t

t = 1.32x10⁻⁷s

<h3>It takes 1.32x10⁻⁷s for the concentration of A to fall by a factor of 8</h3>

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Explanation:

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Sulfur dioxide, SO 2 ( g ) , can react with oxygen to produce sulfur trioxide, SO 3 ( g ) , by the reaction 2 SO 2 ( g ) + O 2 (
aleksley [76]

<u>Answer:</u> The amount of heat produced by the reaction is -21.36 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]

For the given chemical reaction:

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(SO_3(g))})]-[(2\times \Delta H_f_{(SO_2(g))})+(1\times \Delta H_f_{(O_2(g))})]

We are given:

\Delta H_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H_f_{(SO_3(g))}=-395.7kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(2\times (-395.7))]-[(2\times (-296.8))+(1\times (0))]\\\\\Delta H_{rxn}=-197.8kJ/mol

To calculate the number of moles, we use ideal gas equation, which is:

PV=nRT

where,

P = pressure of the gas = 1.00 bar

V = Volume of the gas = 2.67 L

n = number of moles of gas = ?

R = Gas constant = 0.0831\text{ L. bar }mol^{-1}K^{-1}

T = temperature of the mixture = 25^oC=[25+273]K=298K

Putting values in above equation, we get:

1.00bar\times 2.67L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1\times 2.67}{0.0831\times 298}=0.108mol

To calculate the heat released of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat released = ?

n = number of moles = 0.108 moles

\Delta H_{rxn} = enthalpy change of the reaction = -197.8 kJ/mol

Putting values in above equation, we get:

-197.8kJ/mol=\frac{q}{0.108mol}\\\\q=(-197.8kJ/mol\times 0.108mol)=-21.36kJ

Hence, the amount of heat produced by the reaction is -21.36 kJ

3 0
3 years ago
1. How many milliliters of 10.0 M HNO 3 are needed to prepare 0.350 L of 0.400 M solution?
tatyana61 [14]
<h3>Answer:</h3>

14 milliliters

<h3>Explanation:</h3>

We are given;

  • 10.0 M HNO₃

Prepared solution;

  • Volume of solution as 0.350 L
  • Molarity as 0.40 M

We are required to determine the initial volume of HNO₃

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M₁V₁ = M₂V₂

Rearranging the formula;

V₁ = M₂V₂ ÷ M₁

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But, 1 L = 1000 mL

Therefore,

Volume = 14 mL

Thus, the volume of 10.0 M HNO₃ is 14 mL

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Answer:

A. 6N

B. 4H, 2O

C. 4H, 4N, 12O

D. 2Ca, 4O, 4H

E. 3Ba, 6Cl, 18O

F. 5Fe, 10N, 30O

G. 12Mg, 8P, 32O

H. 4N, 16H, 2S, 8O

I. 12Al, 18Se, 72O

J. 12C, 32H

I am 90% sure this is correct

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