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laiz [17]
4 years ago
5

The unit for density found in many density tables is kg/m3. at a certain temperature, the gasoline you pump into your car’s gas

tank has a density of 732.22 kg/m3. if your tank has a capacity of 13 gallons, how many grams of gasoline are in your tank when it is full? how many pounds?
G?
Lbs?
Chemistry
1 answer:
kaheart [24]4 years ago
6 0
To determine the mass of gasoline in the the tank, we need data on the density of gasoline. We multiply this value to the capacity of the gasoline tank. We do as follows:

Mass = 13 gallons (0.0038 m^3 / 1 gallons) ( 732.22 kg/m^3 ) = 36.17 kg or 361771.67 g

36.17 kg ( 2.2 lb / 1 kg ) = 79.58 lb
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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
hjlf

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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