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mr Goodwill [35]
3 years ago
7

What is the systematic name of Mg(NO3)2?

Chemistry
1 answer:
goldenfox [79]3 years ago
5 0
Mg(NO3)2 => <span>Magnesium nitrate

hope this helps!</span>
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Which type of fat is a solid at room temperature and does not have double bonds between carbon
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Also, Saturated fat is very unhealthy. Only consume it in small amounts.

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6.285×10^3 mg = _____? _____ kg
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5.11 g of MgSO₄ is placed into 100.0 mL of water. The water's temperature increases by 6.70°C. Calculate ∆H, in kJ/mol, for the
cupoosta [38]

Answer: Thus ∆H, in kJ/mol, for the dissolution of MgSO₄ is -66.7 kJ

Explanation:

To calculate the entalpy, we use the equation:

q=mc\Delta T

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q = heat absorbed by water = ?

m = mass of water = {\text {volume of water}}\times {\text {density of water}}=100.0ml\times 1.00g/ml=100.0g

c = heat capacity of water = 4.186 J/g°C

\Delta T= change in temperature = 6.70^0C

q=100.0g\times 4.184J/g^0C\times 6.70^0C=2803.3J=2.8033kJ

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

The heat absorbed by water will be equal to heat released by MgSO_4

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass = 5.11 g

Molar mass  = 120 g/mol

Putting values in above equation, we get:

\text{Moles of }MgSO_4=\frac{5.11g}{120g/mol}=0.042mol

0.042 moles of MgSO_4 releases = 2.8033 kJ

1 mole of MgSO_4 releases = \frac{2.8033 kJ}{0.042}\times 1=66.7kJ

Thus ∆H, in kJ/mol, for the dissolution of MgSO₄ is -66.7 kJ

3 0
3 years ago
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