Answer:
c. 0.750 atm
.
Explanation:
Hello!
In this case, since the two vessels have different volume, we can see that the gas is initially at 3.00 atm into the 1.00-L vessel, but next, it is allowed to move towards the 3.00-L vessel, meaning that the final volume wherein the gas is located, is 4.00 L; therefore, we use the Boyle's law to compute the final pressure:
Therefore the answer is c. 0.750 atm
.
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Consisting of parts all of the same kind material.
Answer:
Blue
Explanation:
If you look at a flame, blue is always at the bottom right? So that would be common sense that blue would be the hottest.
<span>STP means standard temperature
and pressure at 0°C (273K) and 1 atm (atmosphere). The density of the unknown
gas is 0.63 gram per liter. The deal gas equation is PV = nRT. The n is the
numer of moles and can be represented as mass of the gas, m, divided by the
molar mass, c. so we have,</span>
PV = nRT
PV = (m/c)RT
Since the density is d = m/V
Pc = (m/V)RT
Pc = dRT
c = drT/P
substitute the values into the equation,
c = [(0.63g/L)(0.08206
L-atm/mol-K)(273K)]/(1atm)
<u>c = 14.11 g/mol</u>
Answer:
The vapor pressure of benzaldehyde at 61.5 °C is 70691.73 torr.
Explanation:
- To solve this problem, we use Clausius Clapeyron equation: ln(P₁/P₂) = (ΔHvap / R) (1/T₁ - 1/T₂).
- The first case: P₁ = 1 atm = 760 torr and T₁ = 451.0 K.
- The second case: P₂ = <em>??? needed to be calculated</em> and T₂ = 61.5 °C = 334.5 K.
- ΔHvap = 48.8 KJ/mole = 48.8 x 10³ J/mole and R = 8.314 J/mole.K.
- Now, ln(P₁/P₂) = (ΔHvap / R) (1/T₁ - 1/T₂)
- ln(760 torr /P₂) = (48.8 x 10³ J/mole / 8.314 J/mole.K) (1/451 K - 1/334.5 K)
- ln(760 torr /P₂) = (5869.62) (-7.722 x 10⁻⁴) = -4.53.
- (760 torr /P₂) = 0.01075
- Then, P₂ = (760 torr) / (0.01075) = 70691.73 torr.
So, The vapor pressure of benzaldehyde at 61.5 °C is 70691.73 torr.
