Answer:
The composition of air this is because it vmade up of oxygen, nitrogen, Nobel gages and Carbon dioxide
Answer:
1.584e10 = 15,840,000,000,000
Explanation:
250,000 miles
multiply the length value by 63360
25e4 x 63360
= 1.584e10
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:
7
Explanation:
Assume we have 1 L of each solution.
Solution 1
![\text{[H$^{+}$]}= 10^\text{-pH} \text{ mol/L} = 10^{\text{-2}} \text{ mol/L}\\ \text{ moles of H}^{+} = \text{ 1 L solution} \times \dfrac{10^{-2}\text{ mol H}^{+}}{\text{1 L solution}} = 10^{-2}\text{ mol H}^{+}](https://tex.z-dn.net/?f=%5Ctext%7B%5BH%24%5E%7B%2B%7D%24%5D%7D%3D%2010%5E%5Ctext%7B-pH%7D%20%5Ctext%7B%20mol%2FL%7D%20%3D%2010%5E%7B%5Ctext%7B-2%7D%7D%20%5Ctext%7B%20mol%2FL%7D%5C%5C%20%5Ctext%7B%20moles%20of%20H%7D%5E%7B%2B%7D%20%3D%20%5Ctext%7B%201%20L%20solution%7D%20%5Ctimes%20%5Cdfrac%7B10%5E%7B-2%7D%5Ctext%7B%20mol%20H%7D%5E%7B%2B%7D%7D%7B%5Ctext%7B1%20L%20solution%7D%7D%20%3D%2010%5E%7B-2%7D%5Ctext%7B%20mol%20H%7D%5E%7B%2B%7D)
Solution 2
pH = 12
pOH = 14.00 - pOH = 14.00 - 12 = 2.0
![\text{[OH$^{-}$]}= 10^\text{-pOH} \text{ mol/L} = 10^{\text{-2}} \text{ mol/L}\\ \text{ moles of OH}^{-} = \text{ 1 L solution} \times \dfrac{10^{-2}\text{ mol OH}^{-}}{\text{1 L solution}} = 10^{-2}\text{ mol OH}^{-}](https://tex.z-dn.net/?f=%5Ctext%7B%5BOH%24%5E%7B-%7D%24%5D%7D%3D%2010%5E%5Ctext%7B-pOH%7D%20%5Ctext%7B%20mol%2FL%7D%20%3D%2010%5E%7B%5Ctext%7B-2%7D%7D%20%5Ctext%7B%20mol%2FL%7D%5C%5C%20%5Ctext%7B%20moles%20of%20OH%7D%5E%7B-%7D%20%3D%20%5Ctext%7B%201%20L%20solution%7D%20%5Ctimes%20%5Cdfrac%7B10%5E%7B-2%7D%5Ctext%7B%20mol%20OH%7D%5E%7B-%7D%7D%7B%5Ctext%7B1%20L%20solution%7D%7D%20%3D%2010%5E%7B-2%7D%5Ctext%7B%20mol%20OH%7D%5E%7B-%7D)
3. pH after mixing
H⁺ + OH⁻ ⟶ H₂O
I/mol: 10⁻² 10⁻²
C/mol: -10⁻² -10⁻²
E/mol: 0 0
The H⁺ and OH⁻ have neutralized each other. The pH will be that of pure water.
pH = 7