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3 years ago

What is the amount of aluminum chloride produced from 40 moles of chlorine and excess aluminum?

Chemistry

1 answer:

Vera_Pavlovna [14]3 years ago

3 0

Answer: Molar mass of Al = 26.98 g/mol

mass of Al = 12 g

mol of Al = (mass)/(molar mass)

= 12/26.98

= 0.4448 mol

According to balanced equation

mol of AlCl3 formed = moles of Al

= 0.4448 mol

Answer: 0.445 mol

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Explanation:

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Answer:

$E=2052.8 kJ$

Explanation:

The energy absorbed by the water is the energy it requires to evaporate. So:

$Heat of .vap= 40.65 kJ/mol$

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$n_w=m_w *\frac{1000g}{2.2lbs}*\frac{1}{M}$

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$n_w=2 lbs *\frac{1000g}{2.2lbs}*\frac{1}{18g/mol}$

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The energy absorbed:

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3 years ago

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A sample of He gas (3.0 L) at 5.6 atm and 25°C was combined with 4.5 L of Ne gas at 3.6 atm and 25°C at constant temperature in

In-s [12.5K]

Answer:

3.6667

Explanation:

For helium gas:

Using Boyle's law

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Using above equation as:

${P_1}\times {V_1}={P_2}\times {V_2}$

${5.6}\times {3.0}={P_2}\times {9.0} atm$

${P_2}=\frac {{5.6}\times {3.0}}{9.0} atm$

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The pressure exerted by the helium gas in 9.0 L flask is 1.8667 atm

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Using Boyle's law

${P_1}\times {V_1}={P_2}\times {V_2}$

Given ,

V₁ = 4.5 L

V₂ = 9.0 L

P₁ = 3.6 atm

P₂ = ?

Using above equation as:

${P_1}\times {V_1}={P_2}\times {V_2}$

${3.6}\times {4.5}={P_2}\times {9.0} atm$

${P_2}=\frac {{3.6}\times {4.5}}{9.0} atm$

${P_1}=1.8\ atm$

The pressure exerted by the neon gas in 9.0 L flask is 1.8 atm

Thus total pressure = 1.8667 + 1.8 atm = 3.6667 atm.

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3 years ago

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Answer:

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Explanation:

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Answer:

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Explanation:

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