Hello!
To do this, use the molar mass. This is how much a mole of an atom weighs. A mole is 6.02214076×10²³ atoms.
Molar masses of:
Se: 78.96 g/mol
Cu: 63.546 g/mol
Ba: 137.327 g/mol
Now, the element with the highest molar mass will have the fewest atoms. This is because the element weighs more, so therefore for the same amount of mass, there will be less of the element needed to reach that mass.
Therefore, 10g of Ba would have the fewest number of atoms.
Hope this helps!
Answer:
The partial pressure of Ar is 356.04 mm Hg (= 0.4685 atm)
Explanation:
<u>Step 1:</u> Data given
A mixture of three gases has a total pressure of 1380 mm Hg (=1.81579 atm) at 298 K
Moles of CO2 = 1.27 moles
Moles of CO = 3.04 moles
Moles of Ar = 1.50 moles
<u>Step 2:</u> Calculate total number of moles
Total number of moles = n(CO2)+ n(CO)+ n(Ar) = 1.27 mol+ 3.04 mol+ 1.50 mol = 5.81 moles
<u>Step 3:</u> Calculate mol fraction Ar
Mol fraction Ar = 1.50 mol/5.81 mol = 0.258
<u>Step 4</u>: Calculate partial pressure
1380 mm Hg * 0.258 moles Ar = 356.04 mm Hg = 0.4685 atm
The partial pressure of Ar is 356.04 mm Hg (= 0.4685 atm)
Answer:
The partial pressure of argon in the jar is 0.944 kilopascal.
Explanation:
Step 1: Data given
Volume of the jar of air = 25.0 L
Number of moles argon = 0.0104 moles
Temperature = 273 K
Step 2: Calculate the pressure of argon with the ideal gas law
p*V = nRT
p = (nRT)/V
⇒ with n = the number of moles of argon = 0.0104 moles
⇒ with R = the gas constant = 0.0821 L*atm/mol*K
⇒ with T = the temperature = 273 K
⇒ with V = the volume of the jar = 25.0 L
p = (0.0104 * 0.0821 * 273)/25.0
p = 0.00932 atm
1 atm =101.3 kPa
0.00932 atm = 101.3 * 0.00932 = 0.944 kPa
The partial pressure of argon in the jar is 0.944 kilopascal.
Water is the only one of these that would work by process of elimination.