First, we need to get n1 (no.of moles of water ): when
mass of water = 0.0203 g and the volume = 1.39 L
∴ n1 = mass / molar mass of water
= 0.0203g / 18 g/mol
= 0.00113 moles
then we need to get n2 (no of moles of water) after the mass has changed:
when the mass of water = 0.146 g
n2 = mass / molar mass
= 0.146g / 18 g/ mol
= 0.008 moles
so by using the ideal gas formula and when the volume is not changed:
So, P1/n1 = P2/n2
when we have P1 = 1.02 atm
and n1= 0.00113 moles
and n2 = 0.008 moles
so we solve for P2 and get the pressure
∴P2 = P1*n2 / n1
=1.02 atm *0.008 moles / 0.00113 moles
= 7.22 atm
∴the new pressure will be 7.22 atm
Answer:
25.35%
Explanation:
Again let me restate the the equation of the reaction;
H2O (ℓ) + 2 MnO4 - (aq) + 3 CN- (aq) → 2 MnO2 (s) + 3 CNO- (aq) + 2 OH- (aq)
Amount of potassium permanganate reacted = 10.2/1000 * 0.08035 = 8.1957 * 10^-4 moles
If 2 moles of MnO4 - reacts with 3 moles of CN-
8.1957 * 10^-4 moles of MnO4 - reacts with 8.1957 * 10^-4 * 3/2
= 1.229 * 10^-3 moles of CN-
Mass of CN- reacted = 1.229 * 10^-3 moles of CN- * 26.02 g/mol
= 0.03 g
Hence, percentage of the cyanide = 0.03 g/0.1183 g * 100
= 25.35%
Answer:
5 L
Explanation:
Use Charles law and rearrange formula
Change C to K
- Hope that helped! Please let me know if you need further explanation.
