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.
To find the mass of glucose, you must multiply the atomic weight of each of the elements in the molecule by the subscripts in the formula:
Then you add all of them together:
Therefore, the molar weight of glucose is 180.15 grams.
<h3>
Answer:</h3>
2.47 × 10^24 molecules
<h3>
Explanation:</h3>
One mole of a compound contains molecules equivalent to the Avogadro's number, 6.022 × 10^23.
That is, 1 mole of a compound = 6.022 × 10^23 molecules
Therefore,
1 mole of Na₂CO₃ = 6.022 × 10^23 molecules
Thus, we can calculate the number of molecules in 4.1 moles of Na₂CO₃
we get,
= 4.1 moles × 6.022 × 10^23 molecules
= 2.47 × 10^24 molecules
Hence, 4.1 moles of Na₂CO₃ contains 2.47 × 10^24 molecules
Answer:The new volume is 5mL
Explanation:
The formular for Boyles Law is; P1 V1 = P2 V2
Where P1 = 1st Pressure V1 = First Volume
P2 = 2nd Pressure V2 = Second Volume
From the question; P1 = 5atm, V1 = 10ml
P2 = 2 x P1 (2 x 5) = 10 atm V2 =?
Using the Boyles Law Formular; P1 V1 = P2 V2, we make V2 the subject of formular; P1 V1/ P2 = V2
∴ 5 x 10/ 10 = 5
∴ V2 = 5mL
