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Semenov [28]
3 years ago
10

Calculate the percentage of water of crystallisation in MgSO⁴ 7H²O

Chemistry
2 answers:
Bas_tet [7]3 years ago
7 0

Answer:

Formula of EPSOM salt = MgSO4.7H2O

molecular mass of MgSO4.7H2O = atomic mass of Mg + atomic mass of S + 4 × atomic mass of O + 7 { 2 × atomic mass of H + atomic mass of O }

= 24 + 32 + 4× 16 + 7{ 2 × 1 + 16 } g/mol

= (24 + 32 + 64+ 126 ) g/mol

= 246 g/mol

molecular mass of total water = 7 × ( 2× atomic mass of H + atomic mass of O )

= 7 × 18 = 126 g/mol

now ,

% mass of H2O in EPSOM salt = {total molar mass of H2O/molar mass of Epsom salts }× 100

= {126/246 } × 100

= 12600/246

= 51.21 %

Explanation:

i have done it hope it helps

lyudmila [28]3 years ago
6 0
I think that to my calculations it should equal to 51.20 I am not sure but I tried x
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In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char
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Answer : The partial pressure of N_2,H_2\text{ and }NH_3 at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

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Initial pressure of H_2 = 2.87 bar

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The given equilibrium reaction is,

                              N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)

Initially                   1.42      2.87             0

At equilibrium    (1.42-x)  (2.87-3x)     2x

The expression of K_p will be,

K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}

Now put all the values of partial pressure, we get

0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}

By solving the term x, we get

x=0.287\text{ and }3.889

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of NH_3 at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of N_2 at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of H_2 at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of N_2 + Partial pressure of H_2 + Partial pressure of NH_3

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

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