The solution for the question above is:
C = 0.270
<span>V = 0.0275L </span>
<span>n = ? </span>
<span>Use the molar formula which is: C = n/V </span>
<span>Re-arrange it to: n = CV </span>
<span>n = (0.270)*(0.0275) </span>
<span>n = 0.007425 mols </span>
<span>(more precise) n = 7.425 x 10^-3 mols
</span>
7.425 x 10^-3 mols is the answer.
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
D IS THE ANSWER TO YOUR QUESTION
Your Answer Will Be Intensive Property
Balance Chemical Equation for combustion of Propane is as follow,
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
According to equation,
1 mole of C₃H₈ on combustion gives = 4 moles of H₂O
So,
5 moles of C₃H₈ on combustion will give = X moles of H₂O
Solving for X,
X = (5 mol × 4 mol) ÷ 1 mole
X = 20 moles of H₂O
Calculating number of molecules for 20 moles of H₂O,
As,
1 mole of H₂O contains = 6.022 × 10²³ molecules
So,
20 moles of H₂O will contain = X molecules
Solving for X,
X = (20 mole × 6.022 × 10²³ molecules) ÷ 1 mol
X = 1.20 ×10²⁵ Molecules of H₂O