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Anarel [89]
3 years ago
15

Calculate the heat required to melt 7.35 g of benzene at its normal melting point. Heat of fusion (benzene) = 9.92 kJ/mol Heat =

kJ
Calculate the heat required to vaporize 7.35 g of benzene at its normal boiling point. Heat of vaporization (benzene) = 30.7 kJ/mol Heat = kJ
Chemistry
1 answer:
grigory [225]3 years ago
3 0

Answer:

The heat required to melt 7.35 g of benzene at its normal melting point is 934.8 Joules.

The heat required to vaporize 7.35 g of benzene at its normal melting point is 2,893 Joules.

Explanation:

Mass of benzene = 7.35 g

Moles of benzene = \frac{7.35 g}{78 g/mol}=0.09423 mol

Heat fusion of benzene,\Delta H_{fus} = 9.92 kJ/mol

1) Heat required to melt 7.35 g of benzene at its normal melting point = Q

Q=\Delta H_{fus}\times 0.09423 mol

=9.92 kJ/mol\times 0.09423 mol=0.9348 kJ=934.8 J

(1 kJ = 1000 J)

2) Heat vaporization of benzene,\Delta H_{vap} = 30.7 kJ/mol

Heat required to vaporize 7.35 g of benzene at its normal melting point = Q

Q=\Delta H_{Vap}\times 0.09423 mol

=30.7 kJ/mol\times 0.09423 mol=2.893 kJ=2,993 J

(1 kJ = 1000 J)

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Answer:

23.8g of sodium phosphate are formed

Explanation:

Based on the reaction of sodium, Na, with phosphoric acid, H₃PO₄:

3Na + H₃PO₄ → Na₃PO₄ + 3/2 H₂

<em>3 moles of sodium produce 1 mole of sodium phosphate</em>

<em />

To solve this question we must find the moles of sodium in 10g. Using the chemical reaction we can find the moles -And the mass- of sodium phosphate produced, as follows:

<em>Moles Na -Molar mass: 22.99g/mol-</em>

10g * (1mol / 22.99g) = 0.435 moles Na

<em>Moles Na₃PO₄:</em>

0.435 moles Na * (1mol Na₃PO₄ / 3mol Na) = 0.145 moles Na₃PO₄

<em>Mass Na₃PO₄ -Molar mass: 163.94g/mol-</em>

0.145 moles Na₃PO₄ * (163.94g/mol) =

<h3>23.8g of sodium phosphate are formed</h3>
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2 years ago
The kinetic energy of a moving object depends on its mass and its
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I think the answer is a, volume, but I still might be wrong.
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Is the process represented below an example of a physical or a chemical
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A mixture of calcium carbonate, CaCO3, and barium carbonate, BaCO3, weighing 5.40 g reacts fully with hydrochloric acid, HCl(aq)
amid [387]

Answer:

CaCO₃ = 85.18%

BaCO₃ = 14.82%

Explanation:

The acid will react with the salts, the partial reactions are:

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

BaCO₃ + 2HCl → BaCl₂ + CO₂ + H₂O

So, the total amount of CaCO₃ BaCO₃ will form the CO₂.

Using the ideal gas law to calculate the number of moles of CO₂:

PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.082 atm*L/mol*K), and T is the temperature (50ºC + 273 = 323 K).

0.904*1.39 = n*0.082*323

26.486n = 1.25656

n = 0.05 mol

So, the number of moles of the mixture is 0.05 mol.

The molar masses of the components are:

CaCO₃ = 40 g/mol of Ca + 12 g/mol of C + 3*16 g/mol of O = 100 g/mol

BaCO₃ = 137.3 g/mol of Ba + 12 g/mol of C + 3*16 g/mol of O = 197.3 g/mol

Let's call x the number of moles of CaCO₃ and y the number of moles of BaCO₃, so:

100x + 197.3y = 5.4

x + y = 0.05 mol

y = 0.05 - x

100x + 197.3*(0.05 - x) = 5.4

100x - 197.3x = 5.4 - 9.865

97.3x = 4.465

x = 0.046 mol of CaCO₃

y = 0.004 mol of BaCO₃

So, the masses are:

CaCO₃ = 100* 0.046 = 4.60 g

BaCO₃ = 137.3*0.004 = 0.80 g

The percentages in the mixture are:

CaCO₃ = (4.60/5.40)*100% = 85.18%

BaCO₃ = (0.80/5.40)*100% = 14.82%

4 0
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. Calculate the number of moles in 56g of H20
svetoff [14.1K]
<h3>Answer:</h3>

3.11 moles

<h3>Explanation:</h3>

We are given 56 g of water (H₂O)

Required to calculate the number of moles of water.

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in this case;

  • Mass of water = 56 g
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Therefore;

Moles of water = 56 g ÷ 18.02 g/mol

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Therefore, moles of water in 56 g will be 3.11 moles

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