Answer:
CH₂ ; 67.1 %
Explanation:
To determine the empirical formula we need to find what the mole ratio is in whole numbers of the atoms in the compound. To do that we will first need the atomic weights of C and H and then perform our calculation
Assume 100 grams of the compound.
# mol C = 85.7 g / 12.01 g/mol = 7.14 mol
# mol H = 14.3 g / 1.008 g/mol = 14.19 mol
The proportion is 14.9 mol H/ 7.14 mol C = 2 mol H/ 1 mol C
So the empirical formula is CH₂
For the second part we will need to first calculate the theoretical yield for the 12.03 g NaBH₄ reacted and then calculate the percent yield given the 0.295 g B₂H₆ produced.
We need to calculate the moles of NaBH₄ ( M.W = 37.83 g/mol )
1.203 g NaBH₄ / 37.83 g/mol = 0.0318 mol
Theoretical yield from balanced chemical equation:
0.0318 mol NaBH₄ x 1 mol B₂H₆ / mol NaBH₄ = 0.0159 mol B₂H₆
Theoretical mass yield B₂H₆ = 0.0159 mol x 27.66 g/ mol = 0.440 g
% yield = 0.295 g/ 0.440 g x 100 = 67.1 %
Boiling water, steam from a cup of tea and ice melting
Answer is: Kb for methylamine is 4.37·10⁻⁴.<span>
Chemical reaction: CH</span>₃NH₂ + H₂O → CH₃NH₃⁺ + OH⁻.
c(CH₃NH₂) = 0.253 M.
α = 4.07% ÷ 100% = 0.0407.
[CH₃NH₃⁺] = [OH⁻] = c(CH₃NH₂) · α.
[CH₃NH₃⁺] = [OH⁻] = 0.253 M · 0.0407.
[CH₃NH₃⁺] = [OH⁻] = 0.0103 M.
[CH₃NH₂] = 0.253 M - 0.0103 M.
[CH₃NH₂] = 0.2427 M.
Kb = [CH₃NH₃⁺] · [OH⁻] / [CH₃NH₂].
Kb = (0.0103 M)² / 0.2427 M.
Kb = 4.37·10⁻⁴.
HCl + NaOH → NaCl + H2O
No. of moles of NaOH = MV/1000
= 3.2X60/1000
= 0.192 mol
**M is your concentration of solution, V is the volume of solution either in cm³ or mL cuz they are equal**
1 mol of NaOH reacted with 1 mol of HCl. No. of moles of HCl needed to react with 0.192mol of NaOH
= 0.192x1/1
= 0.192mol
Volume of HCl = no. of moles x 1000 / M
= 0.192 x 1000 / 4.0
= 48mL
What is the concentration?
