PH is calculated using <span>Handerson- Hasselbalch equation,
pH = pKa + log [conjugate base] / [acid]
Conjugate Base = Acetate (CH</span>₃COO⁻)
Acid = Acetic acid (CH₃COOH)
So,
pH = pKa + log [acetate] / [acetic acid]
We are having conc. of acid and acetate but missing with pKa,
pKa is calculated as,
pKa = -log Ka
Putting value of Ka,
pKa = -log 1.76 × 10⁻⁵
pKa = 4.75
Now,
Putting all values in eq. 1,
pH = 4.75 + log [0.172] / [0.818]
pH = 4.072
Answer:
2
Explanation:
Data:
[H⁺] = 0.01 mol·L⁻¹
Calculation:
pH = -log[H₃O⁺] = -log(0.01) = -log(1) - log(10⁻²) = -0 - (-2) = 0 + 2 = 2
Let's look at the molar weight of the answers:
NO is 30 g/mol
NO2 is 46
N2O is 44
N2O4 is 124
<span>We have the grams of the product, so we need the moles in order to calculate the molar weight. We us PV=nRT for this, assuming standard temperature and pressure. </span>
You were given the liters (.120L)
Std pressure is 1 atmosphere
You're looking for n, the number of moles
<span>Temp is 293.15 kelvin, thats standard </span>
And r is the gas constant in liters-atm per mol kelvin
(.120 liters)(1atm)=n(293.15K)(.08206)
Solving for n is .0049883835 mol
<span>.23g divided by .0049883 mol is about 46g/mol. You're answer is B I think, NO2
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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Ideal solutions obey Raoult's law, which states that:
P_i = x_i*(P_pure)_i
where
P_i is the partial pressure of component i above a solution
x_i is the mole fraction of component i in the solution
(P_pure)_i is the vapor pressure of pure component i
In this case,
P_benzene = 0.59 * 745 torr = 439.6 torr
P_toluene = (1-0.59) * 290 torr = 118.9 torr
The total vapor pressure above the solution is the sum of the vapor pressures of the individual components:
P_total = (439.6 + 118.9) torr = 558.5 torr
Assuming the gas phase also behaves ideally, the partial pressure of each gas in the vapor phase is proportional to its molar concentration, so the mole fraction of toluene in the vapor phase is:
118.9 torr/558.5 torr = 0.213