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2 years ago

Question 2

Chemistry

1 answer:

azamat2 years ago

8 0

Answer:

Explanation:

For A, 2Al(OH)3 contains 6H on the left side of the equation but only has 3H on the right side of the equation, thus its not balanced

For B, there is 2(Al2), which is 4, on the left side of the equation but only 2Al on the right side of the equation, so it is not balanced.

For D, theres a total of 6 H on the left side of the equation as there is (OH)3 and 3H, but none of the right side, so its not balanced.

C is the correct answer as

on the left side of the equation:

2 Al(OH)3 has 2Al, 6O, and 6H

on the right side of the equation:

Al2O3 has 2Al and 3O

3 H20 has 3H and 6O

this makes a total of 2Al, 6H and 6O, which is equal to the number of each element in the left side, thus it is balanced and is the right answer.

In the future for these type of qns, you want to add up the number of each element on the left, then the ride, and see compare the numver of element they have. if the number of all the elements on both sides are the same, the equation is balanced.

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Answer:

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3 years ago

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Answer:

Explanation:

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2 years ago

Read 2 more answers

Consider an amphoteric hydroxide, m(oh)2(s), where m is a generic metal. estimate the solubility of m(oh)2 in a solution buffere

Colt1911 [192]

Missing data in your question: (please check the attached photo)
from this balanced equation:
M(OH)2(s) ↔ M2+(aq) + 2OH-(aq) and when we have Ksp = 2x10^-16
∴Ksp = [M2+][OH]^2
2x10^-16 = [M2+][OH]^2
a) SO at PH = 7
∴POH = 14-PH = 14- 7 = 7
when POH = -㏒[OH]
7= -㏒[OH]
∴[OH] = 1x10^-7 m by substitution with this value in the Ksp formula,
∴[M2+] =Ksp /[OH]^2
= (2x10^-16)/(1x10^-7)^2
= 0.02 M
b) at PH =10
when POH = 14- PH = 14-10 = 4
when POH = -㏒[OH-]
    4 = -㏒[OH-]
∴[OH] = 1x10^-4 ,by substitution with this value in the Ksp formula
[M2+] = Ksp/ [OH]^2
= 2x10^-16 / (1x10^-4)^2
  = 2x10^-8 M
c) at PH= 14
when POH = 14-PH
= 14 - 14
= 0
when POH = -㏒[OH]
  0 = - ㏒[OH]
∴[OH] = 1 m
by substitution with this value in Ksp formula :
[M2+] = Ksp / [OH]^2
= (2x10^-16) / 1^2
= 2x10^-16 M

8 0

3 years ago