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2 years ago

How much energy is required to vaporize 1.5 KG of aluminum

Chemistry

1 answer:

Soloha48 [4]2 years ago

5 0

Answer:

The amount of energy required to vaporize 1.5 kg of aluminium is 16.345 GJ

Explanation:

Given that the heat of vaporization of aluminium, ΔHvap = 294000 kJ/mol

Mass of the given aluminium = 1.5 kG = 1500 g

Number of moles of aluminium = Mass of aluminium/(Molar Mass of aluminium)

Molar Mass of aluminium = 26.98 g/mol

Therefore, we have;

Number of moles = 1500/26.98 = 55.6 moles

The energy required = The heat of vaporization of aluminium × Number of moles of aluminium

The energy required = 294000 × 55.6 = 16345441.0675 kJ = 16.345 GJ.

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The volume of a gas at 7.00°c is 49.0 ml. if the volume increases to 74.0 ml and the pressure is constant, what will the tempera

marshall27 [118]

Using charles law
v1/t1=v2/t2
v1=49ml
v2=74
t1=7+273=280k
t2=?
49/280=74/t2
0.175=74/t2 cross multiply
0.175t2=74
t2=74/0.175
t2=422k or 149celcius

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2 years ago

How does the equilibrium change to counter the removal of A in this reaction? A + B ⇌ AB

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To counter the removal of A the equilibrium change by shifting toward the left

explanation

If the reaction is at equilibrium and we alter the condition a new equilibrium state is created

The removal of A led to the shift of equilibrium toward the left since it led to less molecules in reactant side which favor the backward reaction.( equilibrium shift to the left)

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2 years ago

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Consider the following balanced thermochemical equation for a reaction sometimes used for H2S production:

fgiga [73]

Answer:

d. Heat is released from the reaction

Explanation:

A negative enthalpy change indicates that it is an exothermic reaction. Exothermic reactions release heat.

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2 years ago

Perform the following calculations and give your answer with the correct number of significant figures:

Arte-miy333 [17]

When multiplying numbers, the term with the least significant digits gives how many significant digits will be in the answer. 2.995 has the least with 4 "sig figs", so the answer will have 4 significant digits as well:

2.995/0.16685 = 17.9502547198
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4 sig figs

So the answer is 17.95.

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2 years ago

Acetonitrile (CH3CN) is a polar organic solvent that dissolves a wide range of solutes, including many salts. The density of a 1

erastovalidia [21]

Answer:

a. [LiBr] = 2.70 m

b. Xm for LiBr = 0.1

c. 81% by mass CH₃CN

Explanation:

Solvent → Acetonitrile (CH₃CN)

Solute → LiBr, lithium bromide

We convert the moles of solute to mass → 1.80 mol . 86.84 g/1 mol = 156.3 g

This mass of solute is contained in 1L of solution

1 L = 1000 mL → 1mL = 1cm³

We determine solution mass by density

Solution density = Solution mass / Solution volume

Solution density . Solution volume = solution mass

0.824 g/cm³ . 1000 cm³ = 824 g

Mass of solution = 824 g (solvent + solute)

Mass of solute = 156.3 g

Mass of solvent = 824 g - 156.3 g = 667.7 g

Molality → Moles of solute in 1kg of solvent

We convert the mass of solvent from g to kg → 667.7 g . 1kg /1000g = 0.667 kg

Mol/kg → 1.80 mol / 0.667 kg = 2.70 m → molality

Mole fraction → Mole of solute / Total moles (moles solute + moles solvent)

Moles of solvent → 667.7 g . 1mol/ 41g = 16.3 moles

Total moles = 16.3 + 1.8 = 18.1

Mole fraction Li Br → 1.80 moles / 18.1 moles = 0.1

Mass percentage → (Mass of solvent, in this case / Total mass) . 100

We were asked for the acetonitrile → (667.7 g / 824 g) . 100 = 81%

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2 years ago