Answer:
V = 14.2 L
Explanation:
Given data:
Moles of CO₂ = 0.632 mol
Temperature = standard = 273 K
Pressure = standard = 1 atm
Volume of gas = ?
Solution;
Formula:
PV = nRT
R = general gas constant = 0.0821 atm.L/ mol.K
Now we will put the values in formula.
V = nRT/P
V = 0.632 mol ×0.0821 atm.L/ mol.K × 273 K / 1 atm
V = 14.2 L/ 1
V = 14.2 L
a balanced chemical equation occurs when the number of the atoms involved in the reactants side is equal to the number of atoms in the products side.
Answer:
the work done on the gas is 4,988.7 J.
Explanation:
Given;
number of moles of the monoatomic gas, n = 4 moles
initial temperature of the gas, T₁ = 300 K
final temperature of the gas, T₂ = 400 K
The work done on the gas is calculated as;
For monoatomic ideal gas: 
Where;
R is ideal gas constant = 8.3145 J/K.mol
Therefore, the work done on the gas is 4,988.7 J.
Answer:
V = 331.13 mL of 50 by mass NO3 solution
Explanation:
∴ δ sln = 2.00 g/mL
∴ %m HNO3 = 50% = ( mass HNO3 / mass sln ) × 100
∴ V required sln = 500 mL
∴ δ HNO3 = 1.51 g/mL.......from literature
⇒ V sln = mass sln / δ sln = 500 mL
⇒ mass sln = (500 mL )×( 2 g/mL ) = 1000 g sln
⇒ mass HNO3 = ( 0.5 )×(1000 g) = 500 g HNO3
⇒ V HNO3 = ( 500 g HNO3 )×(ml HNO3/1.51 g ) = 331.13 mL
