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3 years ago

Sr 32.8 g, Si 5.2 g and O 11.9 g empirical formula number 11

Chemistry

1 answer:

horsena [70]3 years ago

8 0

$\boxed{\begin{array}{c|c|c|c|c}\boxed{\sf Element}&\boxed{\sf Mass\:in\: compound}&\boxed{\sf No\:of\:moles}&\boxed{\sf Ratio}&\boxed{\sf Simplified\:ratio}\\ \sf Sr &\sf 32.8g &\sf \dfrac{32.8}{87}=0.37&\sf \dfrac{0.37}{0.18}=2.05&\sf 2 \\ \sf Si &\sf 5.2g &\sf \dfrac{5.2}{28}=0.18 &\sf \dfrac{0.18}{0.18}=1&\sf 1\\ \sf O&\sf 11.9g &\sf \dfrac{11.9}{16}=0.74&\sf\dfrac{0.74}{0.18}=4.1 &\sf 4\end{array}}$

Empirical formula:-

$\\ \sf\longmapsto SiSr_2O_4$

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3 years ago

Given the data calculated in Parts A, B, C, and D, determine the initial rate for a reaction that starts with 0.85 M of reagent

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Answer : The initial rate for a reaction will be $3.8\times 10^{-4}Ms^{-1}$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The chemical equation will be:

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Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b[C]^c$

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a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

$6.1\times 10^{-5}=k(0.2)^a(0.2)^b(0.2)^c$ ....(1)

Expression for rate law for second observation:

$1.8\times 10^{-4}=k(0.2)^a(0.2)^b(0.6)^c$ ....(2)

Expression for rate law for third observation:

$2.4\times 10^{-4}=k(0.4)^a(0.2)^b(0.2)^c$ ....(3)

Expression for rate law for fourth observation:

$2.4\times 10^{-4}=k(0.4)^a(0.4)^b(0.2)^c$ ....(4)

Dividing 1 from 2, we get:

$\frac{1.8\times 10^{-4}}{6.1\times 10^{-5}}=\frac{k(0.2)^a(0.2)^b(0.6)^c}{k(0.2)^a(0.2)^b(0.2)^c}\\\\3=3^c\\c=1$

Dividing 1 from 3, we get:

$\frac{2.4\times 10^{-4}}{6.1\times 10^{-5}}=\frac{k(0.4)^a(0.2)^b(0.2)^c}{k(0.2)^a(0.2)^b(0.2)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.4)^a(0.4)^b(0.2)^c}{k(0.4)^a(0.2)^b(0.2)^c}\\\\1=2^b\\b=0$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^0[C]^1$

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$6.1\times 10^{-5}=k(0.2)^2(0.2)^0(0.2)^1$

$k=7.6\times 10^{-3}M^{-2}s^{-1}$

Now we have to calculate the initial rate for a reaction that starts with 0.85 M of reagent A and 0.70 M of reagents B and C.

$\text{Rate}=k[A]^2[B]^0[C]^1$

$\text{Rate}=(7.6\times 10^{-3})\times (0.85)^2(0.70)^0(0.70)^1$

$\text{Rate}=3.8\times 10^{-3}Ms^{-1}$

Therefore, the initial rate for a reaction will be $3.8\times 10^{-3}Ms^{-1}$

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