Answer:
1.15 atm
Explanation:
According to Dalton's law of partial pressures, the total pressure is the sum of all the partial pressures of the gases present in the mixture.
Therefore we have:
Total pressure = partial pressure of carbon monoxide + partial pressure of oxygen + partial pressure of carbon dioxide
We were given the following:
Total pressure = 2.45 atm
Pressure of oxygen = 0.65 atm
Pressure of carbon monoxide = x
Pressure of carbon dioxide = 0.65 atm
Therefore:
2.45 = x + 0.65 + 0.65
2.45 = x + 1.3
x = 2.45 - 1.3
x = 1.15 atm
I attached a photo of my work below. Please tell me if I got it wrong; hope you got a good score on your assignment! ^v^
Answer is:
7.8 lb of 21% aluminum and 33.2 ib of <span>
42% aluminum.</span>
ω₁<span> = 21% ÷ 100% = 0.21.
ω</span>₂<span> = 42% ÷ 100% = 0.42.
ω</span>₃<span> = 38% ÷ 100% = 0.38.
</span>m₁ = ?.
m₂<span> = ?.
</span>m₃ = m₁ + m₂<span>.
</span>m₃ = 41 pounds.
m₁ = 41 lb - m₂<span>.
ω</span>₁ · m₁ + ω₂ ·m₂ = ω₃ · m₃.
0.21 · (41 lb -
m₂) + 0.42 · m₂ = 0.38 · 41 lb.
8.61 lb - 0.21m₂ + 0.42m₂ = 15.58 lb.
0.21m₂ = 6.97 lb.
m₂ = 6.97 lb ÷ 0.21.
m₂ = 33.2 lb.
m₁ = 41 lb - 33.2 lb.
m₁<span> = 7.8 lb.</span>
Answer:
![[H^{+}] = 0.761 \frac{mol}{L}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%200.761%20%5Cfrac%7Bmol%7D%7BL%7D)
![[OH^{-}]=1.33X10^{-14}\frac{mol}{L}](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D%3D1.33X10%5E%7B-14%7D%5Cfrac%7Bmol%7D%7BL%7D)
Explanation:
HCl and HNO₃ both dissociate completely in water. A simple method is to determine the number of moles of proton from both these acids and dividing it by the total volume of solution.
. V_{HCl}(L) \\ n_{H^{+} } from HNO_{3} = [HNO_{3}](\frac{mol}{L}). V_{HNO_{3}}(L)](https://tex.z-dn.net/?f=n_%7BH%5E%7B%2B%7D%20%7D%20from%20HCl%20%3D%20%5BHCl%5D%28%5Cfrac%7Bmol%7D%7BL%7D%29.%20V_%7BHCl%7D%28L%29%20%20%5C%5C%20n_%7BH%5E%7B%2B%7D%20%7D%20from%20HNO_%7B3%7D%20%20%3D%20%5BHNO_%7B3%7D%5D%28%5Cfrac%7Bmol%7D%7BL%7D%29.%20V_%7BHNO_%7B3%7D%7D%28L%29)
Here, n is the number of moles and V is the volume. From the given data moles can be calculated as follows
For molar concentration of hydrogen ions:
![[H^{+}] = \frac{n_{H^{+}}(mol)}{V(L)}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%20%3D%20%5Cfrac%7Bn_%7BH%5E%7B%2B%7D%7D%28mol%29%7D%7BV%28L%29%7D)
![[H^{+}] = \frac{0.761}{1.00}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%20%5Cfrac%7B0.761%7D%7B1.00%7D)
From dissociation of water (Kw = 1.01 X 10⁻¹⁴ at 25°C) [OH⁻] can be determined as follows
![K_{w} = [H^{+} ][OH^{-} ]](https://tex.z-dn.net/?f=K_%7Bw%7D%20%3D%20%5BH%5E%7B%2B%7D%20%5D%5BOH%5E%7B-%7D%20%5D)
![[OH^{-}]=\frac{Kw}{[H^{+}] }](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D%3D%5Cfrac%7BKw%7D%7B%5BH%5E%7B%2B%7D%5D%20%7D)
![[OH^{-}]=\frac{1.01X10-^{-14}}{0.761 }](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D%3D%5Cfrac%7B1.01X10-%5E%7B-14%7D%7D%7B0.761%20%7D)
The pH of the solution can be measured by the following formula:
![pH = -log[H^{+} ]](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH%5E%7B%2B%7D%20%5D)
CH3-CH=CH2+Br2--->CH3-CH(Br)-CH2-Br
