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777dan777 [17]
3 years ago
7

Urgent! Please help me solve this Chemistry problem.

Chemistry
1 answer:
puteri [66]3 years ago
5 0

Considering the equation,

3Na₂SO₄(aq)+2Al(NO₃)₃(aq)------>Al₂(SO₄)₃(s)+6NaNO₃(aq)

As it can be seen from the equation,

2 moles of Al(NO₃)₃ reacts with 3 moles of Na₂SO₄ to give 1 mol of Al₂(SO₄)₃.

Here 690 mL of 2.5 M Al(NO₃)₃=2.5× 690/1000

=1.7 mol of Al(NO₃)₃

As 2 moles of Al(NO₃)₃ give 1 mol of Al₂(SO₄)₃

So 1.7 mol of Al(NO₃)₃ give 1.7/2 mol of Al₂(SO₄)₃

so it will give 0.86 mol of Al₂(SO₄)₃.

Moles=mass given /molar mass

molar mass of Al₂(SO₄)₃= 342.1509 g/mol

Moles= mass given/342.1509

0.86=mass given/342.1509

mass of Al₂(SO₄)₃=295.1 g

so 295.1 g of Al₂(SO₄)₃ will be produced.

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OLEGan [10]

Answer:

7.21 × 10⁴ J

Explanation:

Ethanol is solid below -114.5°c, liquid between -114.5°C and 78.4°C, and gaseous above 78.4°C.

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<em />

We need to calculate the heat required in different stages and then add them.

The moles of ethanol are:

48.3g.\frac{1mol}{46.07g} =1.05mol

Solid-liquid transition

Q₁ = ΔHfus . n = (4.60 kJ/mol) . 1.05 mol = 4.83 kJ = 4.83 × 10³ J

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ΔHfus: molar heat of fusion

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Q₂ = c(l) . m . ΔT = (2.45 J/g.°C) . 48.3g . [78.4°C-(-114.5°C)] = 2.28 × 10⁴ J

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Q₃ = ΔHvap . n = (38.56 kJ/mol) . 1.05 mol = 40.5 kJ = 40.5 × 10³ J

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Q₄ = c(g) . m . ΔT = (1.43 J/g.°C) . 48.3g . (135.3°C-78.4°C) = 3.93 × 10³ J

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c(g): specific heat capacity of the gas

Total heat required

Q₁ + Q₂ + Q₃ + Q₄ = 4.83 × 10³ J + 2.28 × 10⁴ J + 40.5 × 10³ J + 3.93 × 10³ J = 7.21 × 10⁴ J

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