I love you too funny I love you too
It’s deceleration hope that helps!
Answer:
- The molarity of the student's sodium hydroxide solution is 0.0219 M
Explanation:
<u>1) Chemical reaction.</u>
a) Kind of reaction: neutralization
b) General form: acid + base → salt + water
c) Word equation:
- sodium hydroxide + oxalic acid → sodium oxalate + water
d) Chemical equation:
- NaOH + H₂C₂O₄ → Na₂C₂O₄ + H₂O
b) Balanced chemical equation:
- 2NaOH + H₂C₂O₄ → Na₂C₂O₄ + 2H₂O
<u>2) Mole ratio</u>
- 2mol Na OH : 1 mol H₂C₂O₄ :1 mol Na₂C₂O₄ : 2 mol H₂O
<u>3) Starting amount of oxalic acid</u>
- mass = 28 mg = 0.028 g
- molar mass = 90.03 g/mol
- Convert mass in grams to number of moles, n:
n = mass in grams / molar mass = 0.028 g / 90.03 g/mol = 0.000311 mol
<u>4) Titration</u>
- Volume of base: 28.4 mL = 0.0248 liter
- Concentration of base: x (unknwon)
- Number of moles of acid: 2.52 mol (calculated above)
- Proportion using the theoretical mole ratio (2mol Na OH : 1 mol H₂C₂O₄)

That means that there are 0.000622 moles of NaOH (solute)
<u>5) Molarity of NaOH solution</u>
- M = n / V (liter) = 0.000622 mol / 0.0284 liter = 0.0219 M
That is the correct number using <em>three signficant figures</em>, such as the starting data are reported.
Answer: New concentration of
is 0.23 M.
Explanation:
The given data is as follows.
Moles of
= 0.06 mol
Moles of
= 0.08 mol
Therefore, moles of
added are as follows.
Moles of
= 
= 0.003125 mol
Now, new moles of
= 0.06 + 0.003125
= 0.063125
Therefore, new concentration of
will be calculated as follows.
Concentration = 
= 0.23 M
Thus, we can conclude that new concentration of
is 0.23 M.