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Arisa [49]
3 years ago
6

A 124-mL bubble of hot gas initially at 212 °C and 1.80atm is emitted from an active volcano. What is the final temperature, in

degrees Celcius, of the gas in the bubble outside the volcano when the final volume of the bubble is 138 mL and the pressure is 0.800 atm, if the amount of gas does not change?
Chemistry
1 answer:
romanna [79]3 years ago
4 0

Answer:

H

Explanation:

G

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At a particular temperature, an equilibrium mixture the reaction below was found to contain 0.171 atm of I2, 0.166 atm of Cl2 an
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Answer: 3390

Explanation:

Since this problem already gives is the equilibrium values, all we have to do is to plug them into the formula for K_{p}.

K_{p} =\frac{[ICl]^2}{[I_{2}][Cl_{2}]  }

K_{p} =\frac{(9.81)^2}{(0.171)(0.166)} =3390

7 0
3 years ago
Carbon tetrachloride can be produced by the following reaction: Suppose 1.20 mol of and 3.60 mol of were placed in a 1.00-L flas
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The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)

Suppose 1.20 mol CS_2(g) of and 3.60 mol of Cl_2(g)  were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol  of CCl_4. Calculate equilibrium constant at the unknown temperature.

Answer: The equilibrium constant at unknown temperature is 0.36

Explanation:

Moles of  CS_2 = 1.20 mole

Moles of  Cl_2 = 3.60 mole

Volume of solution = 1.00  L

Initial concentration of CS_2 = \frac{moles}{volume}=\frac{1.20mol}{1L}=1.20M

Initial concentration of Cl_2 = \frac{moles}{volume}=\frac{3.60mol}{1L}=3.60M

The given balanced equilibrium reaction is,

                 CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)

Initial conc.         1.20 M        3.60 M                  0                  0

At eqm. conc.     (1.20-x) M   (3.60-3x) M   (x) M        (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}

Now put all the given values in this expression, we get :

K_c=\frac{(x)\times (x)}{(3.60-3x)^3\times (1.20-x)}

Given :Equilibrium concentration of CCl_4 , x = \frac{moles}{volume}=\frac{0.72mol}{1L}=0.72M

K_c=\frac{(0.72)\times (0.72)}{(3.60-3\times 0.72)^3\times (1.20-0.72)}

K_c=0.36

Thus equilibrium constant at unknown temperature is 0.36

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3 years ago
What volume of oxygen at STP will be produced if the following reaction absorbed 275 kJ of heat?
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Cu + HNO3 g Cu(NO3)2 + NO + H2O
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