Answer:
187.34 atm
Explanation:
From the question,
PV = nRT.................. Equation 1
Where P = Pressure, V = Volume, n = number of mole, R = molar gas constant, T = Temperature.
make P the subject of the equation
P = nRT/V.............. Equation 2
n = mass(m)/molar mass(m')
n = m/m'............... Equation 3
Substitute equation 3 into equation 2
P = (m/m')RT/V............ Equation 4
Given: m = 46 g, T = 25°C = (25+273) = 298 K, V = 3.00 L
Constant: m' = 2 g/mol, R = 0.082 atmL/K.mol
Substitute these values into equation 4
P = (46/2)(0.082×298)/3
P = (23×0.082×298)/3
P = 187.34 atm
Answer:
16.5 dm³
Explanation:
Data Given:
no. moles of O₂ = 0.735 moles
volume of O₂ = ?
Solution:
Now
we have to find volume of O₂ gas
Formula used for this purpose
No. of moles = Volume / molar volume
where
molar volume at STP for Oxygen (O₂) = 22.4 dm³/mol
No. of moles O₂ = Volume of O₂ / 22.4 dm³/mol . . . . . .(1)
Put values in equation 1
0.735 = Volume of O₂ / 22.4 dm³/ mol
rearrange above equation
Volume of O₂ = 0.735 x 22.4 dm³/ mol
Volume of O₂ = 16.5 dm³
So,
the volume of O₂ at STP is 16.5 dm³
<h3><u>Answer</u>;</h3>
= 226 Liters of oxygen
<h3><u>Explanation</u>;</h3>
We use the equation;
LiClO4 (s) → 2O2 (g) + LiCl, to get the moles of oxygen;
Moles of LiClO4;
(500 g LiClO4) / (106.3916 g LiClO4/mol)
= 4.6996 moles
Moles of oxygen;
But, for every 1 mol LiClO4, two moles of O2 are produced;
= 9.3992 moles of Oxygen
V = nRT / P
= (9.3992 mol) x (8.3144621 L kPa/K mol) x (21 + 273) K / (101.5 kPa)
= 226 L of oxygen
Answer:
Explanation:
The formula of calcium nitride is Ca₃P₂.
The masses of each element are:
So, there are 61.94 g of P in 182.18 g of Ca₃P₂.
In 500 g of Ca₃P₂:
There are 
in 500.0 g of Ca₃P₂.
