Answer:
MgSO4.7H2O
Explanation:
Let the formula for the hydrated magnesium sulphate be MgSO4.xH2O
Mass of the hydrated salt (MgSO4.xH2O) = 12.845g
Mass of anhydrous salt (MgSO4) = 6.273g
Mass of water molecule(xH2O) = Mass of the hydrated salt — Mass of anhydrous salt = 12.845 — 6.273 = 6.572g
Now,we can obtain the number of mole of water molecule present in the hydrated salt as follows:
Molar Mass of hydrated salt (MgSO4.xH2O) = 24 + 32 + (16x4) + x(2 + 16) = 24 + 32 + 64 + x(18) = 120 + 18x
Mass of xH2O/ Molar Mass of MgSO4.xH2O = Mass of water / mass of hydrated salt
18x/120 + 18x = 6.572/12.845
Cross multiply to express in linear form
18x x 12.845 = 6.572(120 + 18x)
231.21x = 788.64 + 118.296x
Collect like terms
231.21x — 118.296x = 788.64
112.914x = 788.64
Divide both side by 112.914
x = 788.64 /112.914
x = 7
Therefore the formula for the hydrated salt (MgSO4.xH2O) is MgSO4.7H2O
Answer:
In this experiment, different solutions are made by mixing water with different colors and amounts of food coloring. Students should notice that once the water and colors are mixed together, the liquid looks the same throughout. It is a solution—a homogeneous mixture
Explanation:
Answer:
The water lost is 36% of the total mass of the hydrate
Explanation:
<u>Step 1:</u> Data given
Molar mass of CuSO4*5H2O = 250 g/mol
Molar mass of CuSO4 = 160 g/mol
<u>Step 2:</u> Calculate mass of water lost
Mass of water lost = 250 - 160 = 90 grams
<u>Step 3:</u> Calculate % water
% water = (mass water / total mass of hydrate)*100 %
% water = (90 grams / 250 grams )*100% = 36 %
We can control this by the following equation
The hydrate has 5 moles of H2O
5*18. = 90 grams
(90/250)*100% = 36%
(160/250)*100% = 64 %
The water lost is 36% of the total mass of the hydrate
Answer:
CaO + H20 => Ca(OH)2
Explanation:
quick lime ia a oxyde and when it reacts with water it gives hydroxide
