<span>15.4 milligrams
The ideal gas law is
PV = nRT
where
P = pressure of the gas
V = volume of the gas
n = number of moles of gas
R = Ideal gas constant (8.3144598 L*kPa/(K*mol) )
T = absolute temperature.
So let's determine how many moles of gas has been collected.
Converting temperature from C to K
273.15 + 25 = 298.15 K
Converting pressure from mmHg to kPa
753 mmHg * 0.133322387415 kPa/mmHg = 100.3917577 kPa
Taking idea gas equation and solving for n
PV = nRT
PV/RT = n
n = PV/RT
Substituting known values
n = PV/RT
n = (100.3917577 kPa 0.195 L) / (8.3144598 L*kPa/(K*mol) 298.15 K)
n = (19.57639275 L*kPa) / (2478.956189 L*kPa/(mol) )
n = 0.007897031 mol
So we have a total of 0.007897031 moles of gas particles.
Now let's get rid of that percentage that's water vapor. The percentage of water vapor is the vapor pressure of water divided by the total pressure. So
24/753 = 0.03187251
The portion of hydrogen is 1 minus the portion of water vapor. So
1 - 0.03187251 = 0.96812749
So the number of moles of hydrogen is
0.96812749 * 0.007897031 mol = 0.007645332 mol
Now just multiple the number of moles by the molar mass of hydrogen gas. Start with the atomic weight.
Atomic weight hydrogen = 1.00794
Molar mass H2 = 1.00794 * 2 = 2.01588 g/mol
Mass H2 = 2.01588 g/mol * 0.007645332 mol = 0.015412073 g
Rounding to 3 significant figures gives 0.0154 g = 15.4 mg</span>
Answer:
D
Explanation:
a precipitate is formed from a solution
Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.
Explanation :
First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.
As, 207.2 g of lead contains
atoms
So, 5.5 g of lead contains
atoms
and,
As, 118.71 g of lead contains
atoms
So, 94.5 g of lead contains
atoms
Now we have to calculate the percent composition of Pb and Sn in atom.


and,


Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.
Answer:
0.11 mol
Explanation:
<em>This is the chemical formula for acetic acid (the chemical that gives the sharp taste to vinegar): CH₃CO₂H. An analytical chemist has determined by measurements that there are 0.054 moles of oxygen in a sample of acetic acid. How many moles of hydrogen are in the sample?</em>
Step 1: Given data
- Formula of acetic acid: CH₃CO₂H
- Moles of oxygen in the sample of acetic acid: 0.054 moles
Step 2: Establish the appropriate molar ratio
According to the chemical formula of acetic acid, the molar ratio of H to O is 4:2.
Step 3: Calculate the moles of atoms of hydrogen
We will use the theoretical molar ratio for acetic acid.
0.054 mol O × (4 mol H/2 mol O) = 0.11 mol H