Answer:
282.7KPa
Explanation:
Step 1:
Data obtained from the question.
Number of mole of (n) = 1.5 mole
Volume (V) = 13L
Temperature (T) = 22°C = 22 + 273°C = 295K
Pressure (P) =..?
Gas constant (R) = 0.082atm.L/Kmol
Step 2:
Determination of the pressure exerted by the gas.
This can be obtained by using the ideal gas equation as follow:
PV = nRT
P = nRT /V
P = 1.5 x 0.082 x 295 / 13
P = 2.79atm.
Step 3:
Conversion of 2.79atm to KPa.
This is illustrated below:
1 atm = 101.325KPa
Therefore, 2.79atm = 2.79 x 101.325 = 282.7KPa
Therefore, the pressure exerted by the gas in KPa is 282.7KPa
Answer:
333.3mL
Explanation:
Using the formula as follows:
C1V1 = C2V2
Where;
C1 = initial concentration (M)
C2 = final concentration (M)
V1 = initial volume (mL)
V2 = final volume (mL)
According to the information provided in this question,
C1 = 4.00M
C2 = 1.50M
V1 = 125mL
V2 = ?
Using C1V1 = C2V2
4 × 125 = 1.5 × V2
500 = 1.5V2
V2 = 500/1.5
V2 = 333.3mL
Therefore, the CuSO4 solution needs to be diluted to 333.3mL to make 1.50 M solution.
Answer:11.19
Explanation:
And one mole of a hydrogen atom is of 1.008 grams. So, 2 hydrogen moles weighs 2.016 grams. Hence, one mole of water has 2.016 grams of hydrogen mole. Therefore, the percentage composition of hydrogen would be 2.016/18.0152 = 11.19%.
