4.2 hours, do 315 miles / 75 miles per hour
Answer:
1. C + O₂ → CO₂
2. C + CO₂ → 2 CO
3. Fe₂O₃ + 3 CO → 2 Fe + 3 CO₂
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:
Oxygen, it's supposed to have six valenge electrons.
Explanation:
Count the dots on the oxygen atom, you'll see seven, but there's supposed to be six.
The protons of methylene group between the two carbonyl groups in ethylacetoacetate are acidic in nature. When compounds containing such acidic protons are treated with bases the loose proton and form enolates.
In this particular example when ethylacetoacetate is reacted with methyl magnesium bromide, the methyl group abstracts the acidic proton and converts into
methane gas. The enolate when hydrolyzed is again converted into ethylacetoacetate as shown below,