Answer: 0.52849 j /g °C
Explanation:
Given the following :
Mass of metal = 36g
Δ Temperature of metal = (28.4 - 99)°C = - 70.6°C
Mass of water = 70g
Δ in temperature of water = (28.4 - 24.0) = 4.4°C
Heat lost by metal = (heat gained by water + heat gained by calorimeter)
Quantity of heat(q) = mcΔT
Where; m = mass of object ; c = specific heat capacity of object
Heat lost by metal:
- (36 × c × - 70.6) = 2541.6c - - - - (1)
Heta gained by water and calorimeter :
(70 × 4.184 × 4.4) + (12.4 × 4.4) = 1288.672 + 54.56 = 1343.232 - - - - (2)
Equating (1) and (2)
2541.6c = 1343.232
c = 1343.232 / 2541.6
c = 0.52849 j /g °C
When we have this balanced equation for a reaction:
Fe(OH)2(s) ↔ Fe+2 + 2OH-
when Fe(OH)2 give 1 mole of Fe+2 & 2 mol of OH-
so we can assume [Fe+2] = X and [OH-] = 2 X
when Ksp = [Fe+2][OH-]^2
and have Ksp = 4.87x10^-17
[Fe+2]= X
[OH-] = 2X
so by substitution
4.87x10^-17 = X*(2X)^2
∴X^3 = 4.8x10^-17 / 4
∴the molar solubility X = 2.3x10^-6 M
Answer:So, one mole of water has a mass of 16 +1+1 = 18 grams. So, if one mole has a mass of 18 grams, 25 grams would have a mass of 25 grams/ 18 grams per mole or 1.39 moles
