C) They have a large number of rods and small number of cones.
Answer:
C = 9460 Kj
Explanation:
Given data:
Mass of copper = 2kg
Latent heat of vaporization = 4730 Kj/Kg
Energy required to vaporize 2kg copper = ?
Solution:
Equation
Q= mLvap
by putting values,
Q= 2kg × 4730 Kj/Kg
Q = 9460 Kj
The density would be the same for the whole bar as well as one half of the bar. Density is a identity I believe, by this I mean that it stays the same no matter how little or how much of the same substance you have. Since density = mass / volume, half the bar has half of the weight as well as half of the volume of the whole bar, making the density the same.
For example, a block weighs 10 grams and has a volume of 5 ml. the density would be d = 10/5 or, d = 2g/ml
Half of the block weighs 5 grams and has a volume of 2.5 ml. The density is d = 5/2.5, or, d = 2 g/ml.
See, although there are different amounts of the same substance, their density is the same.
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.
Answer:
239.45 K
Explanation:
Ideal gas law formula is P1V1T2=P2V2T1
Rearrange that to get...
T2=T1P2V2/P1V1
Fill in the values and solve.
