Assuming you are talking about the atomic mass of magnesium chloride
M = n/V
M = 1/3.77
M = .265
The formula of the hydrate = CuSO₄• 3H₂O
<h3>Further explanation</h3>
Given
4.175 grams sample CuSO₄• xH₂O
3.120 grams anhydrous compound CuSO₄
Required
The formula
Solution
mass of H₂O driven off :
= 4.175 - 3.12
= 1.055 g
MW CuSO₄ = 159.5 g/mol
MW H₂O = 18 g/mol
mol ratio of CuSO₄ : H₂O :
= 3.12/159.5 : 1.055/18
= 0.01956 : 0.05861
= 1 : 3
2Ca + O2 = 2CaO
First, determine which is the excess reactant
72.5 g Ca (1 mol) =1.8089725036
(40.078 g)
65 g O2 (1 mol) =2.0313769611
(15.999g × 2)
Since the ratio of to O2 is 2:1 in the balanced reaction, divide Ca's molar mass by 2 to get 0.9044862518. this isn't necessary because Ca is already obviously the limiting reactant. therefore, O2 is the excess reactant.
Now do the stoichiometry
72.5 g Ca (1 mol Ca) (1 mol O2)
(40.078 g Ca)(2 mol Ca)(31.998g O2)
=0.0282669621 g of O2 left over
1. 100 C
2. Point B to C is the ices heat capacity
3. During the points D to E the bonds of the water molecules build up enough kinetic energy to break their intermolecular bonds (not intra), which can lead to gas.
4. Between points D and E the energy is being released the energy required is equivalent along the line.
5. Between point E and D the water is converting to water (condensation)
6. Energy is being released 2260 j/g
7. Yes, but only under extreme volumetric pressures
8. D and E or B and C
9. Freezing (the water is also becoming less dense)
10. Melting or if water already, absorbtion of energy
11. released.
