Using the ideal gas law PV =nRTPV=nRT , we find that the pressure will be P =\frac{nRT}{V}P=
V
nRT
. Then, we'll substitute and find the pressure, using T = -25 °C = 248.15 K and R = 0.0821 \frac{atm\cdot L}{mol \cdot K}
mol⋅K
atm⋅L
:
P =\frac{nRT}{V} = \frac{(0.33\,\cancel{mol})(0.0821\frac{atm\cdot \cancel{L}}{\cancel{mol \cdot K}})(248.15\,\cancel{K})}{15.0\,\cancel{L}} = 0.4482\,atmP=
V
nRT
=
15.0
L
(0.33
mol
)(0.0821
mol⋅K
atm⋅
L
)(248.15
K
)
=0.4482atm
In conclusion, the pressure of this gas is P=0.4482 atm.
Reference:
Chang, R. (2010). Chemistry. McGraw-Hill, New York.
Answer:
2.268 L
Explanation:
According to Ideal Gas Law, " The Volume of a given mass of gas is inversely proportional to the applied Pressure and directly proportional to applied temperature, ". Mathematically the initial and final states of gas are given as,
P₁ V₁ / T₁ = P₂ V₂ / T₂ ----------- (1)
Data Given;
P₁ = 0.998 atm
V₁ = 2.1 L
T₁ = 36 °C = 309 K
P₂ = 0.90 atm
T₂ = 28 °C = 301 K
V₂ = ??
Solving equation 1 for P₂,
V₂ = P₁ V₁ T₂ / T₁ P₂
Putting values,
V₂ = (0.998 atm × 2.1 L × 301 K) ÷ (309 K × 0.90 atm)
V₂ = 2.268 L
Answer:
150 g/mol
Explanation:
Let's consider the complete neutralization of a diprotic acid H₂X with NaOH.
H₂X + 2 NaOH → Na₂X + 2 H₂O
40.0 mL of 0.200 M NaOH. were required to reach the endpoint. The reacting moles of NaOH are:
0.0400 L × 0.200 mol/L = 8.00 × 10⁻³ mol
The molar ratio of H₂X to NaOH is 1:2. The reacting moles of H₂X are 1/2 × 8.00 × 10⁻³ mol = 4.00 × 10⁻³ mol.
4.00 × 10⁻³ moles of H₂X have a mass of 0.600 g. The molar mass of H₂X is:
0.600 g/4.00 × 10⁻³ mol = 150 g/mol
