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2 years ago

Write and balance the equation for the decomposition of aluminum chloride into its elements. phase symbols are optional.

Chemistry

2 answers:

lora16 [44]2 years ago

7 0

The 3 and 2 to the right of the components are subscriptions.

Setler [38]2 years ago

5 0

Explanation:

A balanced equation is an equation in which number of molecules on both left and right side are the same or equal.

When aluminium chloride decomposes then it gives aluminium and chlorine gas.

The decomposition reaction will be as follows.

$2AlCl_{3} (s) \rightarrow 2Al(s) + 3Cl_{2}(g)$

Therefore, we can see that number of molecules on the reactant and product side are same. Hence, this equation is balanced.

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Describe what happens between days 15-28 if the egg does not gets fertilized

dolphi86 [110]

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3 years ago

A geologist working in a national park finds a mineral. He is carrying a kit that contains a penny, a hand lens, a piece of tile

Verizon [17]

A penny for the scratch test (low hardness)a hand lens for inspection a peice of tile for a streak test a geologic hammer to test for cleavage of fracture and the pocket knife for the scratch test (high hardness)

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2 years ago

In the molar volume of a gas lab, a student

GREYUIT [131]

Answer:

Explanation:

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2 years ago

How many moles of compound are there in the following?

slavikrds [6]

a) (NH4)2SO4 --- 1 mole of it contains 2 moles of N, 8 moles of H, 1 mole of S, and 4 moles of O.

MM = (2 moles N x 14.0 g/mole) + (8 moles H x 1.01 g/mole) + (1 mole S x 32.1 g/mole) + (4 moles O x 16.0 g/mole) = 132 g/mole.

6.60 g (NH4)2SO4 x (1 mole (NH4)2SO4 / 132 g (NH4)2SO4) = 0.0500 moles (NH4)2SO4

b) The molar mass for Ca(OH)2 = 74.0 g/mole, calculated like (NH4)2SO4 above.

4.5 kg Ca(OH)2 x (1000 g / 1 kg) x (1 mole Ca(OH)2 / 74.0 g Ca(OH)2) = 60.8 moles Ca(OH)2

8 0

2 years ago

A 2.350×10−2 M solution of NaCl in water is at 20.0∘C. The sample was created by dissolving a sample of NaCl in water and then b

sveta [45]

Answer:

Part A: m = 0.02356 mol/kg = 0.02356 m

Part B: Xsolute = 4.243×10⁻⁴

Part C: % m/m = 0.1376%

Part D: ppm = 1,376 ppm

Explanation:

1. Data:

a) M = 2.350×10⁻² M

b) V sol = 1.000 L

c) V H₂O = 994.4 mL = 0.9944 L

d) d H₂O = 0.9982 g/mL

2. Formulae

M = n solute / V sol (L)
m = n solute / Kg solvent
X solute = n solute / N total
% m/m = (mass of solute / mass of solution) × 100
ppm = (mass of solute / mass of solution) × 1,000,000
density = mass in grams / volume in mL

3. Solution

Part A: Calculate the molality of the salt solution.

m = n solute / Kg solvent

i) M = n solute / V sol (L) ⇒ n solute = M × V sol (L)

⇒ n solute = M = 2.350×10⁻² M × 1.000 L = M = 2.350×10⁻² mol

ii) density H₂O = mass H₂O / volume H₂O

⇒ mass H₂O = density H₂O × volume H₂O

⇒ mass H₂O = 0.9982 g/mL × 999.4 mL = 997.6 g

iii) kg H₂O = 997.6 g / (1,000 g/Kg) = 0.9976 kg

iv) m = 2.350×10⁻² mol / 0.9976 kg = 0.02356 mol/kg = 0.02356 m

Part B: Calculate the mole fraction of salt in this solution.

X solute = n solute / N total

i) n solute = 2.350×10⁻² mol

ii) n solvent = n H₂O = mass H₂O in grams/ molar mass H₂O

⇒ 997.6 g / 18.015 g/mol = 55.38 mol

iii) X solute = 2.350×10⁻² mol / 55.38 mol = 4.243×10⁻⁴

Part C: Calculate the concentration of the salt solution in percent by mass.

% m/m = (mass of solute / mass of solution) × 100

i) molar mass = mass in grams / molar mass

⇒ mass of solute = mass of NaCl = n solute × molar mass NaCl

⇒ mass of solute = 2.350×10⁻² mol × 58.44 g/mol = 1.373 g

ii) % m/m = (1.373 g / 997.6 g) × 100 = 0.1376%

Part D: Calculate the concentration of the salt solution in parts per million.

ppm = (mass of solute / mass of solution) × 1,000,000

i) ppm = ( (1.373 g / 997.6 g) × 1,000,000 = 1,376 ppm

5 0

2 years ago