the properties of liquids usally wet, wet shows expiation on heating and contracting on cooling
Under STP condition, the gas has a rule of 22.4 L per mole. And according to the ideal gas law, V1/T1=V2/T2. Under STP, 1.5 mol gas has volume of 33.6 L. So the volume under 22 C is 33.6*295/273=36.3 L.
Answer:
kp= 3.1 x 10^(-2)
Explanation:
To solve this problem we have to write down the reaction and use the ICE table for pressures:
2SO2 + O2 ⇄ 2SO3
Initial 3.4 atm 1.3 atm 0 atm
Change -2x - x + 2x
Equilibrium 3.4 - 2x 1.3 -x 0.52 atm
In order to know the x value:
2x = 0.52
x=(0.52)/2= 0.26
2SO2 + O2 ⇄ 2SO3
Equilibrium 3.4 - 0.52 1.3 - 0.26 0.52 atm
Equilibrium 2.88 atm 1.04 atm 0.52 atm
with the partial pressure in the equilibrium, we can obtain Kp.

Answer:
2.74 percent hydrogen and 97.3 percent chlorine are in 2 g of compound.
The same as initial
Explanation:
HCl → H⁺ + Cl⁻
2.74% H means that in 100 g of compund, 2.74 grams are H
97.3 % Cl means that in 100 g of compound, 97.3 grams are Cl
So how many grams of H and Cl, are in 1 g of compound
100 g ___ 2.74 g are H _____ 97.3 g are Cl
1g _____ 2.74/100 ____ 97.3/100
0.0274 g are H
0.973 g are Cl
In 2 grams we will find
0.0274 g .2 = 0.0548 g of H
0.973 g .2 = 1.946 g of Cl
(grams / total grams) . 100 = %
(0.0548/2 ) .100 = 2.74%
(1.946 /2) .100 = 97.3%
<u>Answer:</u> The new volume of the tank is 4.67 L
<u>Explanation:</u>
To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.
The equation follows:

where,
are the initial pressure, volume and temperature of the gas
are the final pressure, volume and temperature of the gas
We are given:
Putting values in above equation, we get:

Hence, the new volume of the tank is 4.67 L