Answer:
1.332 g.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- At the same T and P and constant V (1.0 L), different gases have the same no. of moles (n):
<em>∴ (n) of CO₂ = (n) of C₂H₆</em>
<em></em>
∵ n = mass/molar mass
<em>∴ (mass/molar mass) of CO₂ = (mass/molar mass) of C₂H₆</em>
mass of CO₂ = 1.95 g, molar mass of CO₂ = 44.01 g/mol.
mass of C₂H₆ = ??? g, molar mass of C₂H₆ = 30.07 g/mol.
<em>∴ mass of C₂H₆ = [(mass/molar mass) of CO₂]*(molar mass) of C₂H₆</em> = [(1.95 g / 44.01 g/mol)] * (30.07 g/mol) =<em> 1.332 g.</em>
<em></em>
Answer:
21.182 g
Explanation:
There are about (6.0)(10^23) atoms in one mole of a substance, so the given sample has about 0.333 mol of Cu.
The atomic mass of Cu is 63.546 g/mol, meaning that the answer is about <u>21.182</u><u> </u><u>g</u>
Answer:
7200 kPa
Explanation:
Applying,
PV/T = P'V'/T'................ Equation 1
Where P = Initial pressure of neon gas, V = Initial volume of neon gas, T = Initial temperature of neon gas, P' = Final pressure of neon gas, V' = Final volume of neon gas, T' = Final Temperature of neon gas
Make P' the subject of the equation
P' = PVT'/V'T.............. Equation 2
Given: P = 900 kPa, V = 8.0 L, T = 300 K, V' = 2.0 L, T' = 600 K
Substitute these values into equation 2
P' = (900×8×600)/(2×300)
P' = 7200 kPa
