Pressure in the submarine when the temperature is changed to 293 K is 108.9 K Pa
Explanation:
Pressure in the submarine = 108.9 kPa
Volume, V = 2.4 * 10^5 L
Pressure, P = 116k Pa
Temperature, T = 312 K
Ideal gas law: PV = nRT or n = PV / RT
So, moles of gas, n =116 KPa * 2.4 * 10 ^5L / 8.314 LK Pa K^-1 *312 K
= 1.073 *10^4 mol
when temperature is changed to 293K,
PV = nRT or P = nRT / V
=1.073 *10^4 mol *8.314 LK Pa mol^-1 K^-1 *293 K / 2.4*10^5L
=108.9 K Pa
Pressure in the submarine when the temperature is changed to 293 K is 108.9 K Pa
Answer:
19.07 g mol^-1
Explanation:
The computation of the molecular mass of the unknown gas is shown below:
As we know that

where,
Diffusion rate of unknown gas = 155 mL/s
CO_2 diffusion rate = 102 mL/s
CO_2 molar mass = 44 g mol^-1
Unknown gas molercualr mass = M_unknown
Now placing these values to the above formula

After solving this, the molecular mass of the unknown gas is
= 19.07 g mol^-1
Answer : The final volume of gas will be, 26.3 mL
Explanation :
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,

where,
= initial pressure of gas = 0.974 atm
= final pressure of gas = 0.993 atm
= initial volume of gas = 27.5 mL
= final volume of gas = ?
= initial temperature of gas = 
= final temperature of gas = 
Now put all the given values in the above equation, we get:


Therefore, the final volume of gas will be, 26.3 mL
Answer:
Explanation:
70% (vol/vol) means
cotnaimns 70 %(vol/vol) 70 ml of isoprapnol is there in 100 ml of Rubbing sold alcohol.
if it is 200 ml then obvouly it has the 70*2 =140 ml of isoproanol required.