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2 years ago

Calculate the percentage of water of crystallisation in MgSO⁴ 7H²O

Chemistry

2 answers:

Bas_tet [7]2 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]2 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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Answer:

The activation energy for this reaction = 23 kJ/mol.

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Where,

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

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$\left(\ln \left(\:\frac{4.8\times \:\:\:10^8}{2.3\times \:\:\:10^8}\right)\right)\:=-\frac{E_a}{8.314\times \:10^{-3}\ kJ/mol.K}\times \:\left(\frac{1}{649.15\ K}-\frac{1}{553.15\ K}\right)$

$E_a=-\frac{10^{-3}\times \:8.314\ln \left(\frac{10^8\times \:4.8}{10^8\times \:2.3}\right)}{-\frac{96}{359077.3225}}\ kJ/mol$

$E_a=-\frac{\frac{8.314\ln \left(\frac{4.8}{2.3}\right)}{1000}}{-\frac{96}{359077.3225}}\ kJ/mol$

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<u>The activation energy for this reaction = 23 kJ/mol.</u>

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Answer:

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