Answer:
Yes, the chemist can determine which compound is in the sample.
Explanation:
In 1 mole of K₂O, the mass of K is 2 × 39.1 g = 78.2 g and the mass of K₂O is 94.2 g. The mass ratio of K to K₂O is 78.2 g / 94.2 g = 0.830.
In 1 mole of K₂O₂, the mass of K is 2 × 39.1 g = 78.2 g and the mass of K₂O₂ is 110.2 g. The mass ratio of K to K₂O₂ is 78.2 g / 110.2 g = 0.710.
If the chemist knows the mass of K and the mass of the sample, he or she must calculate the mass ratio of K to the sample.
- If the ratio is 0.830, the compound is pure K₂O.
- If the ratio is 0.710, the compound is pure K₂O₂.
- If the ratio is not 0.830 or 0.710, the sample is a mixture.
Answer:
The molecular formula of cacodyl is C₄H₁₂As₂.
Explanation:
<u>Let's assume we have 1 mol of cacodyl</u>, in that case we'd have 209.96 g of cacodyl and the<u> following masses of its components</u>:
- 209.96 g * 22.88/100 = 48.04 g C
- 209.96 g * 5.76/100 = 12.09 g H
- 209.96 g * 71.36/100 = 149.83 g As
Now we convert those masses into moles:
- 48.04 g C ÷ 12 g/mol = 4.00 mol C
- 12.09 g H ÷ 1 g/mol = 12.09 mol H
- 149.83 g As ÷ 74.92 g/mol = 2.00 mol As
Those amounts of moles represent the amount of each component in 1 mol of cacodyl, thus, the molecular formula of cacodyl is C₄H₁₂As₂.
Answer:
Explanation:
<u>1) Data:</u>
a) Hypochlorous acid = HClO
b) [HClO} = 0.015
c) pH = 4.64
d) pKa = ?
<u>2) Strategy:</u>
With the pH calculate [H₃O⁺], then use the equilibrium equation to calculate the equilibrium constant, Ka, and finally calculate pKa from the definition.
<u>3) Solution:</u>
a) pH
b) Equilibrium equation: HClO (aq) ⇄ ClO⁻ (aq) + H₃O⁺ (aq)
c) Equilibrium constant: Ka = [ClO⁻] [H₃O⁺] / [HClO]
d) From the stoichiometry: [CLO⁻] = [H₃O⁺] = 2.29 × 10 ⁻⁵ M
e) By substitution: Ka = (2.29 × 10 ⁻⁵ M)² / 0.015M = 3.50 × 10⁻⁸ M
f) By definition: pKa = - log Ka = - log (3.50 × 10 ⁻⁸) = 7.46
the balanced chemical equation for decomposition of HgO is as follows
2HgO --> 2Hg + O₂
stoichiometry of HgO to O₂ is 2:1
number of HgO moles heated are - 3.00 g / 216.59 g/mol = 0.0139 mol
according to stoichiometry of reaction -
number of O₂ moles formed = 0.0139 mol/ 2 = 0.00695 mol
mass of O₂ to be formed - 0.00695 mol x 32.00 g/mol = 0.2224 g
but the actual yield = 0.195 g
percent yield = actual yield / theoretical yield x 100 %
percent yield = 0.195 g / 0.2224 g x 100 % = 87.7 %
answer is 87.7 %
