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>
Answer:
Final Temperature = 36.54 ⁰C
Explanation:
Lets suppose the gas is acting ideally, then according to Charle's Law, "<em>The volume of a fixed mass of gas at constant pressure is directly proportional to the absolute temperature</em>". Mathematically for initial and final states the relation is as follow,
V₁ / T₁ = V₂ / T₂
Data Given;
V₁ = 32 L
T₁ = 10 °C = 283.15 K ∴ K = °C + 273.15
V₂ = 35 L
T₂ = ??
Solving equation for T₂,
T₂ = V₂ × T₁ / V₁
Putting values,
T₂ = (35 L × 283.15 K) ÷ 32 L
T₂ = 309.69 K ∴ ( 36.54 °C )
Result:
As the volume is increased from 32 L to 35 L, therefore, the temperature must have increased from 10 °C to 36.54 °C.
Answer:
1.52atm is the pressure of the gas
Explanation:
To solve this question we must use the general gas law:
PV = nRT
<em>Where P is pressure in atm = Our incognite</em>
<em>V is volume = 50.5L</em>
<em>n are moles of gas = 3.25moles</em>
<em>R is gas constat = 0.082atmL/molK</em>
<em>And T is absolute temperature = 288.6K</em>
To solve pressure:
P = nRT / V
P = 3.25mol*0.082atmL/molK*288.6K / 50.5L
P = 1.52atm is the pressure of the gas
calcium,phosphorus,potassium,and sulfer
