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Georgia [21]
3 years ago
15

How many grams of dry nh4cl need to be added to 2.00 l of a 0.600 m solution of ammonia, nh3, to prepare a buffer solution that

has a ph of 8.74? kb for ammonia is 1.8×10−5?

Chemistry
1 answer:
Brrunno [24]3 years ago
8 0
Nh4cl = 117.165 gram ( solution is attached )

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A sample of 17.0 M concentrated H2SO4 stock solution with a volume of 25.0 cm3 was diluted to final concentration of 5.0 M H2SO4
Art [367]

Answer:

60 cm³ of water

Explanation:

We'll begin by calculating the volume of the diluted solution. This can be obtained as follow:

Concentration of stock solution (C₁) = 17 M

Volume of stock solution (V₁) = 25 cm³

Concentration of diluted solution (C₂) = 5 M

Volume of diluted solution (V₂) =?

C₁V₁ = C₂V₂

17 × 25 = 5 × V₂

425 = 5 × V₂

Divide both side by 5

V₂ = 425 / 5

V₂ = 85 cm³

Thus, the volume of the diluted solution is 85 cm³

Finally, we shall determine the volume of water needed to dilute the solution. This can be obtained as follow:

Volume of stock solution (V₁) = 25 cm³

Volume of diluted solution (V₂) = 85 cm³

Volume of water =?

Volume of water = V₂ – V₁

Volume of water = 85 – 25

Volume of water = 60 cm³

Therefore, 60 cm³ of water is needed to dilute the solution.

6 0
2 years ago
Calculate the ph of a 0.60 m h2so3, solution that has the stepwise dissociation constants ka1 = 1.5 × 10-2 and 1.82 1.06 1.02 2.
Vsevolod [243]
Missing in your question Ka2 =6.3x10^-8
From this reaction:
 H2SO3 + H2O ↔ H3O+  + HSO3-
by using the ICE table :
                H2SO3     ↔    H3O     +    HSO3- 
intial         0.6                     0                  0
change     -X                      +X                +X
Equ         (0.6-X)                  X                   X

when Ka1 = [H3O+][HSO3-]/[H2SO3]
So by substitution:
1.5X10^-2 = (X*X) / (0.6-X) by solving this equation for X
∴ X = 0.088
∴[H2SO3] = 0.6 - 0.088 = 0.512
[HSO3-] = [H3O+] = 0.088

by using the ICE table 2:
                 HSO3-     ↔   H3O     +     SO3-
initial        0.088              0.088              0
change    -X                      +X                   +X
Equ         (0.088-X)          (0.088+X)          X

Ka2= [H3O+] [SO3-] / [HSO3-]
we can assume [HSO3-] =  0.088 as the value of Ka2 is very small
 6.3x10^-8 = (0.088+X)*X / 0.088
X^2 +0.088 X - 5.5x10^-9= 0 by solving this equation for X
∴X= 6.3x10^-8
∴[H3O+] = 0.088 + 6.3x10^-8
               = 0.088 m ( because X is so small)
∴PH= -㏒[H3O+]
       = -㏒ 0.088 = 1.06 
5 0
3 years ago
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Seawater becomes warmer it expands. Heat in the upper layer of the ocean is released quickly into the atmosphere. However, heat absorbed by the deeper layers of the ocean will take much longer to be released and therefore, be stored in the ocean much longer and have significant impacts on future ocean warming.

An increase in freshwater inputs from mountain glaciers, ice sheets, ice caps, and sea ice, as well as other atmospheric and hydrologic cycles due to rising global surface and ocean temperatures
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Water boiling and turning into steam<br><br> is a chemical or physical change?
olga2289 [7]

Answer:

physical change

Explanation:

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3 years ago
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What is the difference between soluble and insoluble
klemol [59]

Soluble means it can be dissolved

Insoluble means it can't be dissolved

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