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2 years ago

8

A 15.33 g sample of magnesium oxide is found to contain 12.67 g of magnesium. What is the percent (by mass) composition of the m

agnesium in the compound?

1 answer:

MAVERICK [17]2 years ago

3 0

12.67 g / 15.33 g x 100 % = 82.64% Magnesium (Mg)

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The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute th

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The question is incomplete, complete question is :

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25°C.

$\frac{K_1}{K_2}=?$

Activation energy of the reaction 1 , $Ea_1$ = 14.0 kJ/mol

Activation energy of the reaction 2, $Ea_1$ = 11.9 kJ/mol

Answer:

0.4284 is the ratio of the rate constants.

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According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

The expression used with catalyst and without catalyst is,

$\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}$

$\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}$

where,

$K_2$ = rate constant reaction -1

$K_1$ = rate constant reaction -2

Activation energy of the reaction 1 , $Ea_1$ = 14.0 kJ/mol = 14,000 J

Activation energy of the reaction 2, $Ea_1$ = 11.9 kJ/mol = 11,900 J

R = gas constant = 8.314 J/ mol K

T = temperature = $25^oC=273+25=298 K$

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

$\frac{K_1}{K_2}=e^{\frac{11,900- 14,000Jl}{8.314 J/mol K\times 298 K}}=2.3340$

0.4284 is the ratio of the rate constants.

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3 years ago

For the simple decomposition reactionAB(g)→ A(g) + B(g)Rate =k[AB]2 and k=0.2 L/mol*s . How long will it takefor [AB] to reach 1

mixer [17]

Answer:

6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.

Explanation:

$Rate = k[AB]^2$

The order of the reaction is 2.

Integrated rate law for second order kinetic is:

$\frac{1}{[A_t]} = \frac{1}{[A]_0}+kt$

Where, $[A_0]$ is the initial concentration = 1.50 mol/L

$[A_t]$ is the final concentration = 1/3 of initial concentration = $\frac{1}{3}\times 1.50\ mol/L$ = 0.5 mol/L

Rate constant, k = 0.2 L/mol*s

Applying in the above equation as:-

$\frac{1}{0.5} = \frac{1}{1.50}+0.2t$

$\frac{1}{1.5}+0.2x=\frac{1}{0.5}$

$t = 6.66\ s$

<u>6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.</u>

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3 years ago