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Akimi4 [234]
4 years ago
13

Provide the expressions for the equilibrium constant Keq: CH4(g) + 2H2SE(g) ⇄ 4H2(g) +CSe2(g)

Chemistry
1 answer:
Sloan [31]4 years ago
3 0

Answer:

Keq = [H₂]⁴[CSe₂]/[CH₄][H₂Se]².

Explanation:

  • The equilibrium constant expression is the ratio of the concentrations of the products over the reactants.

  • For the reaction:

<em>CH₄(g) + 2H₂Se(g) ⇄ 4H₂(g) + CSe₂(g).</em>

<em></em>

<em>∴ Keq = [products]/[reactants] = [H₂]⁴[CSe₂]/[CH₄][H₂Se]².</em>

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closed circuit

Explanation:

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4Cr(s)+3O2(g)→2Cr2O3(s) calculate how many grams of the product form when 21.4 g of O2 completely reacts
weqwewe [10]

Answer:

= 67.79 g

Explanation:

The equation for the reaction is;

4Cr(s)+3O2(g)→2Cr2O3(s)

The mass of O2 is 21.4 g, therefore, we find the number of moles of O2;

moles O2 = 21.4 g / 32 g/mol

                =0.669 moles

Using mole ratio, we get the moles of Cr2O3;

moles Cr2O3 = 0.669 x 2/3

                       =0.446 moles

but molar mass of Cr2O3 is 151.99 g/mol

Hence,

The mass Cr2O3 = 0.446 mol x 151.99 g/mol

                            <u> = 67.79 g </u>

6 0
3 years ago
Describe a method to calculate the average atomic mass of the sample in the previous question using only the atomic masses of li
alexira [117]

Answer:

Explanation:

To calculate their average atomic masses which is otherwise known as the relative atomic mass, we simply multiply the given abundances of the atoms and the given atomic masses.

The abundace is the proportion or percentage or fraction by which each of the isotopes of an element occurs in nature.

This can be expressed below:

        RAM = Σmₙαₙ

where mₙ is the mass of isotope n

           αₙ is the abundance of isotope n

for this problem:

RAM of Li = m₆α₆ + m₇α₇

       m₆ is mass of isotope Li-6

        α₆ is the abundance of isotope Li-6

       m₇ is mass of isotope Li-7

        α₇ is the abundance of isotope Li-7

3 0
3 years ago
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