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:
Volume
Explanation:
Volume is how much of a substance something can hold!
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The isotope that is more abundant, given the data is isotope Li7
<h3>Assumption</h3>
- Let Li6 be isotope A
- Let Li7 be isotope B
<h3>How to determine whiche isotope is more abundant</h3>
- Molar mass of isotope A (Li6) = 6.02 u
- Molar mass of isotope B (Li7) = 7.02 u
- Atomic mass of lithium = 6.94 u
- Abundance of A = A%
- Abundance of B = (100 - A)%
Atomic mass = [(mass of A × A%) / 100] + [(mass of B × B%) / 100]
6.94 = [(6.02 × A%) / 100] + [(7.02 × (100 - A)) / 100]
6.94 = [6.02A% / 100] + [702 - 7.02A% / 100]
6.94 = [6.02A% + 702 - 7.02A%] / 100
Cross multiply
6.02A% + 702 - 7.02A% = 6.94 × 100
6.02A% + 702 - 7.02A% = 694
Collect like terms
6.02A% - 7.02A% = 694 - 702
-A% = -8
A% = 8%
Thus,
Abundance of B = (100 - A)%
Abundance of B = (100 - 8)%
Abundance of B = 92%
SUMMARY
- Abundance of A (Li6) = 8%
- Abundance of B (Li7) = 92%
From the above, isotope Li7 is more abundant.
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844108
One way is to write down all the numbers and add and write down each number.
The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH
<h3>What is Limiting reagent ?</h3>
The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed.
Given chemical equation in balanced form ;
2NaOH(s) + CO₂(g) → Na₂CO₃(s) + H₂O(l).
According to the Chemical equation ;
- The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH
If 44 g CO₂ requires 80 g of NaOH, therefore, 4.4 g CO₂ will require atleast 8 g of NaOH.
But the available quantity is 5 g NaOH. thus, NaOH is the Limiting reagent.
- 6.625 g of Na₂CO₃ are expected to be produced 5.0 g of NaOH and 4.4 g of CO₂ are allowed to react
As 80 g NaOH produces 106 g of Na₂CO₃.
Therefore 5 g NaoH will produce ;
106 / 80 x 5 = 6.625 g
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