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3 years ago

14

What is the capacity to do work known as?

2 answers:

Butoxors [25]3 years ago

6 0

Answer:

Energy

Explanation:

apeks

Jlenok [28]3 years ago

5 0

Energy

Simply Energy.

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Fudgin [204]

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3 years ago

Nitric acid can be produced by the reaction of gaseous nitrogen dioxide with water. 3 no2(g + h2o(? ?? 2 hno3(? + no(g if 538 l

Phantasy [73]

The balanced chemical reaction is written as:

<span>3NO2 + H2O = 2HNO3 + NO

Assuming that the gases in this reaction are ideal gas, then we can use the conversion from L to moles which is 1 mol of ideal gas is equal to 22.4 L. We calculate as follows:

538 L NO2 ( 1 mol / 22.4L ) ( 1 mol NO / 3 mol NO2 ) ( 22.4 L / 1 mol ) = 179.33 L NO is produced</span>

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3 years ago

A baseball a football and a bowling ball are thrown with the same force

Olegator [25]

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Explanation:

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3 years ago

Chemistry table practice

Sonbull [250]

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3 years ago

Iron(III) oxide and hydrogen react to form iron and water, like this: Fe_2O_3(s) + 3H_2(g) rightarrow 2Fe(s) + 3H_2O(g) At a cer

Sergeeva-Olga [200]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The equilibrium constant is $K_c= 2.8*10^{-4}$

Explanation:

From the question we are told that

The chemical reaction equation is

$Fe_{2} O_{3}_{(s)} + 3H_{2}_{(g)} -----> 2Fe_{(s)} + 3H_{2} O_{(g)}$

The voume of the misture is $V_m = 5.4L$

The molar mass of $Fe_{2} O_{3}_{(s)}$ is a constant with value of $M_{Fe_{2} O_{3}_{(s)} } = 160g/mol$

The molar mass of $H_{2}_{(g)}$ is a constant with value of $H_2 = 2g/mol$

The molar mass of $H_{2}O$ is a constant with value of $H_2O = 18g/mol$

Generally the number of moles is mathematically given as

$No \ of \ moles \ = \frac{mass}{molar\ mass}$

For $Fe_{2} O_{3}_{(s)}$

$No \ of\ moles = \frac{3.54}{160}$

$= 0.022125 \ mols$

For $H_{2}$

$No \ of\ moles = \frac{3.63}{2}$

$= 1.815 \ mols$

For $H_{2}O$

$No \ of\ moles = \frac{2.13}{18}$

$= 0.12 \ mols$

Generally the concentration of a compound is mathematicallyrepresented as

$Concentration = \frac{No \ of \ moles }{Volume }$

For $Fe_{2} O_{3}_{(s)}$

$Concentration[Fe_2 O_3] = \frac{0.222125}{5.4}$

$= 4.10*10^{-3}M$

For $H_{2}$

$Concentration[H_2] = \frac{1.815}{5.4}$

$= 0.336M$

For $H_{2}O$

$Concentration [H_2O] = \frac{0.12}{5.4}$

$= 0.022M$

The equilibrium constant is mathematically represented as

$K_c = \frac{[concentration \ of \ product]}{[concentration \ of \ reactant ]}$

Considering $H_2O \ for \ product$

And $H_2 \ for \ reactant$

At equilibrium the

$K_c = \frac{0.022}{0.336}$

$K_c= 2.8*10^{-4}$

3 0

3 years ago