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jarptica [38.1K]
3 years ago
10

Determine the dissociation constants for the following acids. Express the answers in proper scientific notation where appropriat

e
a. acid a has a pka of 6.0 what is its ka?
b. acid b has a pka of 8.60 whats its ka?
c. acid c has a pka of -2.0 whats its ka?

How do you determine?
Chemistry
1 answer:
Tems11 [23]3 years ago
8 0

Explanation:

Ka is the acid dissociation constant. pKa is simply the -log of this constant.

Mathematically;

pKa = - log Ka

Ka = 10^-pKa

a. acid a has a pka of 6.0 what is its ka?

Ka = 10^-6 = 10⁻⁶

b. acid b has a pka of 8.60 whats its ka?

Ka = 10^-8.6 = 2.51e⁻⁹

c. acid c has a pka of -2.0 whats its ka?

Ka = 10^-(-2) = 10^2 = 100

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Answer:

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Explanation:

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17. If 5.00 L of an ideal gas is cooled from 24°C to - 27000
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Answer:

V₂ = 448.44 L

Explanation:

Given data:

Initial volume = 5.00 L

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Final volume = ?

Final temperature = -27000°C

Solution:

Final temperature = -27000 + 273 = 26727 K

Initial temperature = 24 +273 = 297 K

The given problem will be solve through the Charles Law.

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Mathematical expression:

V₁/T₁ = V₂/T₂

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T₁ = Initial temperature

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Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 5.00 L × 26727 K / 297 k

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