An aqueous solution of potassium sulfate exhibits colligative properties. Colligative properties are properties that depends on the concentration of a substance in a solution. These properties are freezing point depression, vapor pressure lowering, osmotic pressure and boiling point elevation. For this problem we use the concept of freezing point depression since we are given the freezing point of the solution. Freezing point depression is as:
ΔT = -k(f) x m x i
-2.24 - 0 = -1.86 x m x 3
<span>m = 0.4014
Thus, the molality of the solution is 0.4014.</span>
Answer:
-133.2 kJ
Explanation:
Let's consider the following balanced equation.
4 KClO₃(s) → 3 KClO₄(s) + KCl(s)
We can calculate the standard Gibbs free energy of the reaction (ΔG°rxn) using the following expression.
ΔG°rxn = 3 mol × ΔG°f(KClO₄(s)) + 1 mol × ΔG°f(KCl(s)) - 4 mol × ΔG°f(KClO₃(s))
ΔG°rxn = 3 mol × (-303.1 kJ/mol) + 1 mol × (-409.1 kJ/mol) - 4 mol × (-296.3 kJ/mol)
ΔG°rxn = -133.2 kJ
Answer: 1709.4 Joules
Explanation:
The quantity of Heat Energy (Q) released on cooling a heated substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Since Q = ?
M = 18.5 grams
Recall that the specific heat capacity of copper C = 0.385 J/g.C
Φ = 285°C - 45°C = 240°C
Then, Q = MCΦ
Q = 18.5grams x 0.385 J/g.C x 240°C
Q = 1709.4 Joules
Thus, 1709.4 Joules is released when copper is cooled.
Explanation:
substance Q could be <em><u>oxygen (O2)</u></em>
substance R could be <em><u>carbon</u></em><em><u> </u></em><em><u>d</u></em><em><u>i</u></em><em><u>o</u></em><em><u>x</u></em><em><u>i</u></em><em><u>d</u></em><em><u>e</u></em><em><u> </u></em><em><u>(</u></em><em><u>C</u></em><em><u>O</u></em><em><u>2</u></em><em><u>)</u></em>
Answer:
2.059524x10^26 if im not wrong
Explanation:
avogadro's number is 6.022x10^23
