The partial atmospheric pressure (atm) of hydrogen in the mixture is 0.59 atm.
<h3>How do we calculate the partial pressure of gas?</h3>
Partial pressure of particular gas will be calculated as:
p = nP, where
- P = total pressure = 748 mmHg
- n is the mole fraction which can be calculated as:
- n = moles of gas / total moles of gas
Moles will be calculated as:
- n = W/M, where
- W = given mass
- M = molar mass
Moles of Hydrogen gas = 2.02g / 2.014g/mol = 1 mole
Moles of Chlorine gas = 35.90g / 70.9g/mol = 0.5 mole
Mole fraction of hydrogen = 1 / (1+0.5) = 0.6
Partial pressure of hydrogen = (0.6)(748) = 448.8 mmHg = 0.59 atm
Hence, required partial atmospheric pressure of hydrogen is 0.59 atm.
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Answer:
pH = 5.54
Explanation:
The pH of a buffer solution is given by the <em>Henderson-Hasselbach (H-H) equation</em>:
- pH = pKa + log
![\frac{[CH_3COO^-]}{[CH_3COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_3COO%5E-%5D%7D%7B%5BCH_3COOH%5D%7D)
For acetic acid, pKa = 4.75.
We <u>calculate the original number of moles for acetic acid and acetate</u>, using the <em>given concentrations and volume</em>:
- CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻
- CH₃COOH ⇒ 0.345 M * 0.250 L = 0.0862 mol CH₃COOH
The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.
Now we use the H-H equation to <u>calculate the new pH</u>, by using the <em>new concentrations</em>:
- pH = 4.75 + log
= 5.54
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A pH scale runs from 1 to 14 with 7 being neutral.
1-6 has base like properties
8-14 has avid line properties
since this solution has a pH scale of 4.... the solution is basic
Answer : The volume of gas will be 29.6 L
Explanation:
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,

where,
= initial pressure of gas = 12 atm
= final pressure of gas = 14 atm
= initial volume of gas = 23 L
= final volume of gas = ?
= initial temperature of gas = 200K
= final temperature of gas = 300K
Now put all the given values in the above equation, we get the final pressure of gas.


Therefore, the new volume of gas will be 29.6 L