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.
He worked with large numbers of plants.
Hi I need Help with this :
A swimming pool has an inlet pipe that can fill the empty pool in 12 hours with the drain is closed and the inlet pipe is closed the drain can empty the full pool in 20 hours. When the pool is empty how long will it take the inlet pipe to fill the pool drain if it is left open ?
Answer:
Molecular formula = C20H30
Explanation:
NB 440mg = 0.44g, 135mg= 0.135g
From the question, moles of CO2= 0.44/44= 0.01mol
Since 1 mol of CO2 contains 1mol of C, it implies mol of C = 0.01
Also from the question, moles of H2O = 0.135/18= 0.0075mole
Since 1 mol of H2O contains 2mol of H, it implies mol of H = 0.0075×2= 0.015 mol of H
To get the empirical formula, divide by smallest number of mole
Mol of C = 0.01/0.01=1
Mol of H = 0.015/0.01= 1.5
Multiply both by 2 to obtain a whole number
Mol of C =1×2 = 2
Mol of H= 1.5×2 = 3
Empirical formula= C2H3
[C2H3] not = 270
[ (2×12) + 3]n = 270
27n = 270
n=10
Molecular formula= [C2H3]10= C20H30