Answer:
Explanation:
In this problem,
we will determine the enthalpy at the end of heat input and we get the condition is the super heated state.
so the entropy change will be due to latent heat at 150 kPa and also due to temperature change at the super heated state.
All the temperature, enthalpy is evaluated with the help of the steam table.
mass is calculated with the help of a specific volume at initial condition.
<u>see image below</u>
The question is incomplete, here is the complete question:
Calculate the pH of a solution prepared by dissolving 0.370 mol of formic acid (HCO₂H) and 0.230 mol of sodium formate (NaCO₂H) in water sufficient to yield 1.00 L of solution. The Ka of formic acid is 1.77 × 10⁻⁴
a) 2.099
b) 10.463
c) 3.546
d) 2.307
e) 3.952
<u>Answer:</u> The pH of the solution is 3.546
<u>Explanation:</u>
We are given:
Moles of formic acid = 0.370 moles
Moles of sodium formate = 0.230 moles
Volume of solution = 1 L
To calculate the molarity of solution, we use the equation:

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
![pH=pK_a+\log(\frac{[salt]}{[acid]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D%29)
![pH=pK_a+\log(\frac{[HCOONa]}{[HCOOH]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5BHCOONa%5D%7D%7B%5BHCOOH%5D%7D%29)
= negative logarithm of acid dissociation constant of formic acid = 3.75
![[HCOOH]=\frac{0.370}{1}](https://tex.z-dn.net/?f=%5BHCOOH%5D%3D%5Cfrac%7B0.370%7D%7B1%7D)
pH = ?
Putting values in above equation, we get:

Hence, the pH of the solution is 3.546
Answer:
A. 32.6 g/mol
Explanation:
First convert the volume of gas to moles using the ratio 1 mol / 22.4 L at STP.
0.070 L • (1 mol / 22.4 L) = 0.00313 mol
Now divide the grams of gas by the moles of gas:
0.102 g / 0.00313 mol = 32.6 g/mol
Answer:
Yes
Explanation:
Is this a question or what?