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3 years ago

There are 43.2 g of carbon to 14.4 g of hydrogen in a sample of methane (CH4). What percent of 37.8 g of methane is carbon?

Chemistry

1 answer:

EastWind [94]3 years ago

8 0

Using the law of constant proportions which says that within the same compound, elements exist in fixed ratios.

Therefore; we can use the ratio of total mass to the mass of carbon, to determine the amount of carbon in another sample.

Mass C / Mass CH4 = Mass C / Mass CH4

43.2 g / 57.6 g = Mass C / 37.8 g

Mass C = 37.8 g × 43.2 g / 57.6 g

= 28.35 g

Hence; the percentage of carbon will be;

=(28.35/ 37.8 )× 100%

= 75 %

Thus; 75% of 37.8 g of methane is carbon

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Complete Question

The rate of a certain reaction is given by the following rate law:

$rate = k [H_2][I_2]$

rate Use this information to answer the questions below.

What is the reaction order in $H_2$ ?

What is the reaction order in $I_2$ ?

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Answer:

The reaction order in $H_2$ is n = 1

The reaction order in $I_2$ is m = 1

The overall reaction order z = 2

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Explanation:

From the question we are told that

The rate law is $rate = k [H_2][I_2]$

The rate of reaction is $rate = 2.0 *10^{4} M /s$

Let the reaction order for $H_2$ be n and for $I_2$ be m

From the given rate law the concentration of $H_2$ is raised to the power of 1 and this is same with $I_2$ so their reaction order is n=m=1

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At $rate = 2.0 *10^{4} M /s$

$2.0*10^{4} = k [H_2] [I_2] ---(1)$

= > $k = \frac{2.0*10^{4}}{[H_2] [I_2] }$

given that the concentration of hydrogen is doubled we have that

$rate = k [2H_2] [I_2] ----(2)$

=> $k = \frac{rate_n }{ [2H_2] [I_2]}$

So equating the two k

$\frac{2.0*10^{4}}{[H_2] [I_2] } = \frac{rate_n }{ [2H_2] [I_2]}$

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$52 .0 = k(1.8)* (0.82)$

$k = \frac{52 .0}{(1.8)* (0.82)}$

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