To convert 78.1 g of water at 0° C to Ice at -57.1°C; we can do it in steps;
1. Water at 0°C to ice at 0°C
The heat of fusion of ice is 334 J/g;
Heat = 78.1 × 334 = 26085.4 Joules
2. Ice at 0°C to -57.1°C
Specific heat of ice is 2.108 J/g
Heat = 78.1 × 2.108 J/g = 164.6348 Joules
Thus the total heat energy released will be; 26085.4 + 164.6348
= 26250.0348 J or 26.250 kJ
Answer:
[H⁺] = 3.16 × 10⁻⁵ mol/L
Explanation:
Given data:
pH of solution = 4.5
Hydrogen ion concentration = ?
Solution;
pH = -log [H⁺]
we will rearrange this formula:
[H⁺] = 10∧-pH
[H⁺] = 10⁻⁴°⁵
[H⁺] = 3.16 × 10⁻⁵ mol/L
We will get the molality from this formula:
Molality = no.of moles of solute / Kg of solvent
So first we need the no.of moles of KNO3 = the mass of KNO3 / molar mass of KNO3
no.of moles of KNO3 = 175 / 101.01 = 1.73 mol
By substitution in the molality formula:
∴ molality = 1.73 / (750/1000) = 2.3 Molal
Answer:
hope it helps you...
Explanation:
acid rain is harmful for crop,and soil.
