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Diano4ka-milaya [45]
1 year ago
9

A mixture of 2.0 mol of CO(g)? and 2 mol of H2O?(g)? was allowed to come to equilibrium in a 1.0L flask at a high temperature. I

f Kc?=4.0, what is the molar concentration of H2(g) in the equilibrium mixture: CO(g) + H2O(g)\rightleftharpoons CO2(g) +H2(g)
Chemistry
1 answer:
Bess [88]1 year ago
7 0

For a mixture of 2.0 mol of CO(g) and 2 mol of H2O(g) was allowed to come to equilibrium in a 1.0L flask at a high temperature. the molar concentration of H2(g)  is mathematically given as

[H2] = x = 1.33 M

<h3>What is the molar concentration of H2(g) in the equilibrium mixture?</h3>

Generally, the equation for the Chemical reaction  is mathematically given as

Kc = [CO2]*[H2]/[CO]*[H2O]

Therefore

4.0 = (1*x)^2/(2-1*x)^2

x = 1.33

In conclusion, the molar concentration

[H2] = x = 1.33 M

Read more about  Chemical reaction

brainly.com/question/16416932

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Hi can u help me pls? I'm totally stuck . The natural source of acidity in rain water is _____.
kykrilka [37]

Answer-The correct option is option d with says all of the above.

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5 0
3 years ago
Suppose an ice cube weighing 36.0 g at a temperature of 10°C is placed in 360 g water at a temperature of 20°C. Calculate the te
Scilla [17]

Answer:

10.44 °C

Explanation:

When the thermal equilibrium is reached, both of the substances have the same final temperature (T). The liquid water will lose heat, and the ice cube will absorb this heat. The temperature of the ice will increase until it reaches 0°C, at this temperature, it will change of phase for liquid, absorbing heat, but without a change in the temperature. Then the temperature will increase until the equilibrium.

By the energy conservation, the total amount of heat must be equal to 0:

Qice + Qmelting + Qliquid1 + Qliquid2 = 0

Liquid 1 is the ice after melting, and liquid 2 the liquid that was already at the flask. When there's a change of temperature:

Q = n*c*ΔT, where n is the number of moles, c is the heat capacity and ΔT is the temperature change (final - initial). The temperature variation in °C is equal in K, so the temperature may be used in °C.

The melting heat is:

Q = n*Hfus, Hfus = 6007 J/mol

The molar mass of the water is 18 g/mol, so the number of moles of the water and the ice are:

nwater = nliquid1 = 360/18 = 20 moles

nice = 36/18 = 2 moles

Qice + Qmelting + Qliquid1 + Qliquid2 = 0

2*38*(0 - (-10)) + 2*6007 + 2*75*(T - 0) + 20*75*(T - 20) = 0

760 + 12014 + 150T + 1500T - 30000 = 0

1650T = 17226

T = 10.44 °C

4 0
3 years ago
G(1)=0 g(n) =g(n-1)+n g(2)=
Aliun [14]
G(2)=2

For this, you can plug in 2 everywhere you see an n. So the equation will read:
g(2)=g(2-1)+2 -> g(2)=g(1)+2. Since we are given g(1)=0, we can plug in 0 where we see g(1). The equation is now. g(2)=0+2. So, g(2)=2.
6 0
3 years ago
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