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4 years ago

What mass of aluminum is produced by the decomposition of 5.0 kg al2o3?

2 answers:

7 0

The mass for of aluminum that is produced by the decomposition of 5.0 Kg Al2O3 is 2647 g or 2.647 Kg

calculation
Write the equation for decomposition of Al2O3

Al2O3 = 2Al + 3 O2

find the moles of Al2O3 = mass/molar mass

convert 5 Kg to g = 5 x1000 = 5000 grams
molar mass of Al2O3 = 27 x2 + 16 x3 = 102 g/mol

moles =5000 g/ 102 g/mol = 49.0196 moles

by use of mole ratio between Al2O3 to Al which is 1:2 the moles of Al = 49.0196 x2 =98.0392 moles

mass of Al = moles x molar mass

= 98.0392 moles x 27g/mol = 2647 grams or 2647/1000 = 2.647 Kg

solong [7]4 years ago

3 0

Answer: 2.7 kg

Explanation:

1) Find the molar mass of Al₂O₃, using the atomic masses of Al and O:


2x27.0g/mol + 3x16.0g/mol = 102.0 g/mol

2) Find the ratio of Al to Al₂O₃:


2x27.0 / 102.0 = 54.0/102.0 = 27.0/51.0 = 9.0/17.0

3) Use proportionality:

9.0 Kg Al / 17.0 kg Al₂O₃ = x / 5.0 kg


⇒ x = 5.0kg Al₂O₃ x (9.0 kg Al/ 17.0 Kg) Al₂O₃

x = 2.65 kg = 2.7 Kg


Al = 2.7 kg Al

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Determine the percent ionization of a 0.225 M solution of benzoic acid. Express your answer using two significant figures.

Assoli18 [71]

Answer:

1.68% is ionized

Explanation:

The Ka of benzoic acid, C₇H₆O₂, is 6.46x10⁻⁵, the equilibrium in water of this acid is:

C₇H₆O₂(aq) + H₂O(l) ⇄ C₇H₅O₂⁻(aq) + H₃O⁺(aq)

Ka = 6.46x10⁻⁵ = [C₇H₅O₂⁻] [H₃O⁺] / [C₇H₆O₂]

Where [] are concentrations in equilibrium

In equilibrium, some 0.225M of the acid will react producing both C₇H₅O₂⁻ and H₃O⁺, the equilibrium concentrations are:

[C₇H₆O₂] = 0.225-X

[C₇H₅O₂⁻] = X

[H₃O⁺] = X

Replacing:

6.46x10⁻⁵ = [X] [X] / [0.225-X]

1.4535x10⁻⁵ - 6.46x10⁻⁵X = X²

1.4535x10⁻⁵ - 6.46x10⁻⁵X - X² = 0

Solving for X:

X = -0.0038. False solution, there is no negative concentrations.

X = 0.00378M. Right solution.

That means percent ionization (100 times Amount of benzoic acid ionized over the initial concentration of the acid) is:

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3 years ago

A material has an ASTM grain size number of 7. Determine the magnification, if the number of grains per square inch observed is:

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Answer:

A) M = 100X

B) M = 36X

C) M = 178.88X

Explanation:

Given data:

ASTM grain size number 7

a) total grain per inch^2 - 64 grain/inch^2

we know that number of grain per square inch is given as

$Nm = 2^{n-1} (\frac{100}{M})^2$

where M is magnification, n is grain size

therefore we have

$64 = 2^{7-1}(\frac{100}{M})^2$

solving for M we get

M = 100 X

B) total grain per inch^2 = 500 grain/inch^2

we know that number of grain per square inch is given as

$Nm = 2^{n-1} (\frac{100}{M})^2$

where M is magnification, n is grain size

therefore we have $500 = 2^{7-1}(\frac{100}{M})^2$

solving for M we get

M = 36 X

C) Total grain per inch^2 = 20 grain/inch^2

we know that number of grain per square inch is given as

$Nm = 2^{n-1} (\frac{100}{M})^2$

where M is magnification, n is grain size

therefore we have $20 = 2^{7-1}(\frac{100}{M})^2$

solving for M we get

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