Answer: The weight/weight % or percent by mass of the solute is 5.41 %.
Explanation:
Mass of the sodium sulfate,w = 9.74 g
Volume of the water = 165 mL
Density of the water = 1 g/mL
Mass of the water =
Mass of the solution, W:
Mass of solute + Mass of solvent =9.47 g + 165 g=174.47 g
The weight/weight % or percent by mass of the solute is 5.41 %.
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I believe the answer is the third option. Hope this helps! Please tell me if I am wrong or if there was an error in my answer... also sorry this answer is late.
No mathematics the most common way of solving an equation will be PEMDAS so the answer for your question will most likely be subtraction.
Answer:
The answer to your question is V2 = 66.7 ml
Explanation:
Data
Volume 1 = V1 = 400 ml
Pressure 1 = P1 = 1 atm
Volume 2 = V2 = ?
Pressure 2 = P2 = 6 atm
Process
1.- To solve this problem use Boyle's law
P1V1 = P2V2
-solve for V2
V2 = P1V1 / P2
-Substitution
V2 = (1)(400) / 6
-Simplification
V2 = 400 / 6
-Result
V2 = 66.7 ml
Answer:
44.91% of Oxygen in Iron (III) hydroxide
Explanation:
To solve this question we must find the molar mass of Fe(OH)3 and the molar mass of the oxygen in this molecule. Percent composition will be:
<em>Molar mass Oxygen / molar mass Fe(OH)3 * 100</em>
<em />
<em>Molar mass Fe(OH)3 and oxygen:</em>
1Fe = 55.845g/mol*1 = 55.845
3O = 16.00g/mol*3 = 48.00 - Molar mass of Oxygen
3H = 1.008g/mol*3 = 3.024
55.845 + 48.00 + 3.024 =
106.869g/mol is molar mass of Fe(OH)3
% Composition of oxygen is:
48.00g/mol / 106.869g/mol * 100 =
<h3>44.91% of Oxygen in Iron (III) hydroxide</h3>
