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:
Бардык белгилүү авиация мыйзамдарына ылайык, аары учуу мүмкүнчүлүгү жок. Канаттары өтө эле кичинекей, семиз денесин чече албайт. Албетте, аары баары бир учат. Себеби аарылар адам мүмкүн эмес деп эсептеген нерсеге маани бербейт.
Explanation:
Канаттары өтө эле кичинекей, семиз денесин чече албайт. Албетте, аары баары бир учат. Себеби аарылар адам мүмкүн эмес деп эсептеген нерсеге маани бербейт.
Answer:
⇒ 3.312 E-13 mol Ni(OH)2 are soluble in 1 L of solution (NaOH)
Explanation:
S S 2S + 0.0218
∴ pKsp Ni(OH)2 = 15.8 = -Log Ksp
⇒ Ksp = 1.585 E-16 = [ Ni2+ ] * [ OH- ]²
∴ pH = 12.34
⇒ pOH = 14- 12.34 = 1.66 = - Log [ OH- ]
⇒ [ OH- ] = 0.0218 M
⇒ Ksp = 1.585 E-16 = S * ( 2S + 0.0218 )²
if we compare the concentration ( 0.0218 ) with the Ksp ( 1.585 E-16 ), then we can neglect the solubility as adding
⇒ 1.585 E-16 = S * ( 0.0218 )²
⇒ 1.585 E-16 = 4.786 E-4 * S
⇒ S = 3.312 E-13 mol/L
∴ soluble moles Ni(OH)2 = 3.312 E-13 mol/L * 1 L Sln = 3.312 E-13 moles
⇒ % S = ( 3.312 E-13 / 0.0218 ) * 100 = 1.513 E-11 %.....we can accept the previous assumption.
Answer:
a) 2.6 mole of O2 per mole of methane
b) 9.78 mole of N2 per mole of methane
c) Mole fraction of H2O = 0.33
d) 6.12 mole of H2O pero mole of methane
Explanation:
The first thing you have to do is the balanced reaction of the methane combustion:
CH4 +2 O2 -->2 H2O + CO2
You calculate the moles of O2 needed to react with 1 mole of methane and add the excess factor
2 x 1.3 = 2.6 moles of O2
For the N2 moles you calculate it with the air composition (21 % O2, 79% N2)
2.6 moles of O2 * 0.79/0.21 = 9.78 moles of N2
With the total stream of air = 12.38 moles you add the humidity factor
12.38 * 1.5 = 18.57 moles So 6.12 are moles of H20 entering per mole of CH4.
To calculate the mole fraction yo divide the moles of water among the moles of the stream:
6.12/18.57 --> 0.33
Answer:
The answer is 4778.4 cm (cubed)
Explanation:
V = (πr(squared)) x (h/3)
V = (π x 13(squared)) x (27/3)
V = 530.9291585 x 9
V = 4778.4
