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algol [13]
3 years ago
11

1. A mixture contains 8.00 g each of O2, CO2, and SO2 at STP. Calculate the volume of this mixture. Which of the gases would exe

rt the greatest pressure and why?
Chemistry
1 answer:
Gekata [30.6K]3 years ago
7 0

Answer:

Explanation:

mole of O₂ = \frac{8}{32}

= .25 moles

mole of CO₂

= \frac{8}{44}

= .1818 moles

moles of SO₂

\frac{8}{64}

= .125 moles

Total moles of gas

= .5568 moles.

total volume of gas mixture

= 22.4 x .5568 liter ( volume of one mole of any gas = 22.4 liter)

= 12.47 liter.

gas will exert partial pressure according to their mole fraction

gas having greatest no of moles in the total mole will have greatest mole fraction so

O₂ will have greatest partial pressure.

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V(CuSO₄·5H₂O) = 1 L.
c(CuSO₄·5H₂O) = 0,30 mol/L.
n(CuSO₄·5H₂O)  = V(CuSO₄·5H₂O) · c(CuSO₄·5H₂O) .
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Write the equilibrium constant expression for the following reaction in terms of concentrations of the components. (Concentratio
timofeeve [1]

<u>Answer:</u> The expression for equilibrium constant in terms of concentration is K_c=[CO_2]

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric coefficients. It is represented by K_{c}

For a general chemical reaction:

aA+bB\rightarrow cC+dD

The K_{c} is written as:

K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}

The concentration of pure solids and pure liquids are taken as 1.

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MgCO_3(s)\rightarrow MgO(s)+CO_2(g)

The expression for K_{c} is:

K_{c}=\frac{[MgO][CO_2]}{[MgCO_3]}

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So, the expression for K_c becomes:

K_{c}=[CO_2]

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