It is an ideal gas therefore we can use the ideal gas equation to solve the problem. The ideal gas equation is expressed as PV = nRT. First, we solve the amount of the gas in moles using the said equation and the first conditions.
(2.0 atm) (5.0 x 10^3 cm^3) = n (82.0575 atm.cm^3/mol.K)(215 K)
n=0.5668 mol
Using the second conditions given, we obtain the new pressure.
P (4.0 x 10^3) = 0.5668 x <span>82.0575 x 265
P= 3.08 atm</span>
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:
Molarity = moles ÷ liters
to get moles of NaBr divide grams of NaBr by its molar mass (mass of Na + mass of Bromine)
Na = 22.989769
Br = 79.904
molar mass of NaBr = 102.893769
6.6g ÷ 102.893769 = 0.064143826 moles of NaBr
0.064143826 moles ÷ 0.60 liters = 0.1069 molar concentration or 11 %
Answer:
It's true :) Hope that helps
Answer: row 1, the volume decreases when the pressure increased
Explanation:
