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son4ous [18]
2 years ago
8

How many grams of aluminum oxide will you need to start with in order to make 6500 g of aluminum hydroxide?

Chemistry
2 answers:
REY [17]2 years ago
8 0

Answer:

56.7 grams of aluminum oxide (AI²O³) are produced.Need a fast expert's response.

kherson [118]2 years ago
3 0

Answer:

Explanation:

2Al(s) + 3 2 O2(g) → Al2O3(s) And given the stoichiometry ...and EXCESS dioxygen gas...we would get 6.25⋅ mol of alumina. the which represents a mass... ...6.25 ⋅ mol ×101.96 ⋅ g ⋅ mol−1 molar mass of alumina ≡ 637.25 ⋅ g.

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In order to calculate the number of neutrons you must subtract the what from the what?​
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3 years ago
Read 2 more answers
The equilibrium constant for the formation of ammonia from nitrogen and hydrogen is 1.6 × 102. what is the form of the equilibri
Nimfa-mama [501]

Answer: The expression for equilibrium constant is \frac{[NH_3]^2}{[H_2]^3[N_2]}

Explanation: Equilibrium constant is the expression which relates the concentration of products and reactants preset at equilibrium at constant temperature. It is represented as k_c

For a general reaction:

aA+bB\rightleftharpoons cC+dD

The equilibrium constant is written as:

k_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}

Chemical reaction for the formation of ammonia is:

N_2+3H_3\rightleftharpoons 2NH_3

k_c=1.6\times 10^2

Expression for k_c is:

k_c=\frac{[NH_3]^2}{[H_2]^3[N_2]}

1.6\times 10^2=\frac{[NH_3]^2}{[H_2]^3[N_2]}

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3 years ago
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