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3 years ago

Which substance is an Arrhenius acid?a. H2b.HCIc. KCId, NH,3

Chemistry

1 answer:

GenaCL600 [577]3 years ago

4 0

Answer:

Hcl

Explanation:

Arrhenius acid - produce H+

Arrhenius base - produce oh -

we can see here in the option B give us H+ which is arrhenius acid

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Of rational function 1 3. + 3 < 0 3x-1 nctions.

pshichka [43]

Answer:

I didn't understand questions properly. Sorry

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3 years ago

Calculate the pH of a solution prepared by mixing: (Show your work for these calculations) pk of acetic acid is 4.75 a. two mole

kupik [55]

Answer:

Explanation:

To calculate pH you need to use Henderson-Hasselbalch formula:

pH = pka + log₁₀ $\frac{[A^-]}{[HA]}$

Where HA is the acid concentration and A⁻ is the conjugate base concentration.

The equilibrium of acetic acid is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka: 4,75

Where CH₃COOH is the acid and CH₃COO⁻ is the conjugate base.

Thus, Henderson-Hasselbalch formula for acetic acid equilibrium is:

pH = 4,75 + log₁₀ $\frac{[CH_{3}COO^-]}{[CH_{3}COOH]}$

a) The pH is:

pH = 4,75 + log₁₀ $\frac{[2 mol]}{[2 mol]}$

pH = 4,75

b) The pH is:

pH = 4,75 + log₁₀ $\frac{[2 mol]}{[1mol]}$

pH = 5,05

I hope it helps!

7 0

4 years ago

7. The equilibrium constant Kc for the reaction H2(g) + I2(g) ⇌ 2 HI(g) is 54.3 at 430°C. At the start of the reaction there are

Juli2301 [7.4K]

Answer:

[H2] = 0.0692 M

[I2] = 0.182 M

[HI] = 0.826 M

Explanation:

Step 1: Data given

Kc = 54.3 at 430 °C

Number of moles hydrogen = 0.714 moles

Number of moles iodine = 0.984 moles

Number of moles HI = 0.886 moles

Volume = 2.40 L

Step 2: The balanced equation

H2 + I2 → 2HI

Step 3: Calculate Q

If we know Q, we know in what direction the reaction will go

Q = [HI]² / [I2][H2]

Q= [n(HI) / V]² /[n(H2)/V][n(I2)/V]

Q =(n(HI)²) /(nH2 *nI2)

Q = 0.886²/(0.714*0.984)

Q =1.117

Q<Kc This means the reaction goes to the right (side of products)

Step 2: Calculate moles at equilibrium

For 1 mol H2 we need 1 mol I2 to produce 2 moles of HI

Moles H2 = 0.714 - X

Moles I2 = 0.984 -X

Moles HI = 0.886 + 2X

Step 3: Define Kc

Kc = [HI]² / [I2][H2]

Kc = [n(HI) / V]² /[n(H2)/V][n(I2)/V]

Kc =(n(HI)²) /(nH2 *nI2)

KC = 54.3 = (0.886+2X)² /((0.714 - X)*(0.984 -X))

X = 0.548

Step 4: Calculate concentrations at the equilibrium

[H2] = (0.714-0.548) / 2.40 = 0.0692 M

[I2] = (0.984 - 0.548) / 2.40 = 0.182 M

[HI] = (0.886+2*0.548) /2.40 = 0.826 M

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3 years ago

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3 years ago

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Which substance is an Arrhenius acid?a. H2b.HCIc. KCId, NH,3​

Which substance is an Arrhenius acid?a. H2b.HCIc. KCId, NH,3