A.
elements C H N
percentage composition 74.03 8.70 17.27
Molecular mass 12 1 14
# of mole 6.17 8.70 1.23
÷smallest mole 5.0 7.0 1.0
mole ratio 5 : 7 ; 1
THE EMPERICAL FORMUKLA FOR A. IS C5H7O
NOTE: #of mole = percentage composition ÷ Mr
and the ÷ smallest mole is used to find the ratio...for the above question a. it is 1.23
and b. should be done using the same procedure
Answer:
2.5 %
Explanation:
Considering:
Or,
Given :
For 
:
Molarity = 0.2850 M
Volume = 63.30 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 63.30 × 10⁻³ L
Thus, moles of :
Moles of = 0.0180405 moles
Moles of = Moles of 
Thus, Moles of = 0.0180405 moles
Molar mass of = 35.453 g/mol
Mass = Moles * Molar mass = 0.0180405 moles * 35.453 g/mol = 0.6396 g
Volume of sea water = 25.00 mL
Density = 1.024 g/mL
Density = Mass / Volume
Mass = Density * Volume = 1.024 g/mL * 25.00 mL = 25.6 g
<u>Mass percent of Cl⁻ = 2.5 %</u>
Explanation:
1) The dissolution of the salt potassium sulfite:
K₂SO₃(aq) → 2K⁺(aq) + SO₃²⁻(aq).
Potassium has +1 charge because it lost one electron to accomplish stabile electron configuration of noble gas argon.
2) From dissolution reaction: n(K⁺) : n(SO₃²⁻) = 2 : 1.
n(K⁺) = 0.700 mol.
0.700 mol : n(SO₃²⁻) = 2 : 1.
n(SO₃²⁻) = 0.700 mol ÷ 2.
n(SO₃²⁻) = 0.350 mol; amount of sulfite anions.
