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SpyIntel [72]
2 years ago
15

Art a the ph of a 0. 25 m aqueous solution of hydrofluoric acid, hf, at 25. 0 °c is 2. 3. What is the value of ka for hf

Chemistry
1 answer:
avanturin [10]2 years ago
8 0

The value of K_{a} for HF is, 3.5 x 10^{-4} M

<h3>What is an aqueous solution?</h3>

The aqueous solution definition means simply that something has been dissolved in water. The aqueous symbol is (aq). Aqueous solution definition: something is dissolved in water.

Solution : Given,

pH = 2.03

Concentration of HF = 0.25 M

First, we have to calculate the concentration of H^{+} ion.

pH=-log [H^{+}]

2.03=-log [H^{+}]

H^{+}= 9.3 x 10^{-3} M

Now we have to calculate the value of  K_{a} for HF.

The equilibrium reaction will be:

HF\leftrightharpoons \ H^{+}+F^{-}

Concentration of  H^{+} = Concentration of  F^{-}= 9.3 x 10^{-3} M

The expression of K_{a} for HF will be,

k_{a} =\frac{[F]^{-} [H]^{+} }{HF}

K_{a} = 3.5 x 10^{-4} M

Therefore, the value of K_{a} for HF is, 3.5 x 10^{-4} M

Learn  more about the aqueous solution here:

brainly.com/question/14097392

#SPJ4

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What type of isomerism is shown by alkanes
Luda [366]

Answer:

Alkanes with more than 3 carbons can show constitutional isomerism. They can be either linear or branched structures. This is categorized as chain isomerism. Butane is the smallest alkane to show such isomerism with 2 isomers.Alkanes with more than 3 carbons can show constitutional isomerism. They can be either linear or branched structures. This is categorized as chain isomerism. Butane is the smallest alkane to show such isomerism with 2 isomers.

Explanation:

6 0
3 years ago
Given: logK=nE∘0.0592 What is the value K for this redox reaction? Zn2+(aq) + 2 Cl−(aq) → Zn(s) + Cl2(
anygoal [31]

<em>K</em> = 2.4 × 10^(-72)

<em>Step 1</em>. Determine the <em>value of n </em>

Zn^(2+) + 2e^(-) → Zn

2Cl^(-) → Cl_2 + 2e^(-)

Zn^(2+) + 2Cl^(-) → Zn + Cl_2

∴ <em>n</em> = 2

<em>Step 2</em>. Calculate <em>K</em>

log<em>K</em> = <em>nE</em>°/0.0592 V = [2 × (-2.12 V)]/0.0592 V = -71.62

<em>K</em> = 10^(-71.62) = 2.4 × 10^(-72)

8 0
3 years ago
8. Sulfur has a first ionization energy of 1000 kJ/mol. Photons of what frequency are required to ionize one mole of Sulfur?​
Lynna [10]

Answer:

the frequency of photons v = 1.509\times10^{39}Hz

Explanation:

Given:  first ionization energy of 1000 kJ/mol.

No. of moles of sulfur = 1 mole

\Delta E_1 = 1000KJ/mol

We know that plank's constant

h = 6.626\times10^{-34} Js

Let the frequency of photons be ν

Also we know that ΔE = hν

this implies ν = ΔE/h

= \frac{10^6J}{6.626\times10^{-34} Js}

v = 1.509\times10^{39}Hz

Hence, the frequency of photons v = 1.509\times10^{39}Hz

6 0
3 years ago
Calculate the Ka for the following acid. Determine if it is a strong or weak acid.
vladimir1956 [14]

Answer:

a) Ka= 7.1 × 10⁻⁴; This is a weak acid because the acid is not completely dissociated in solution.

Explanation:

Step 1: Write the dissociation reaction for nitrous acid

HNO₂(aq) ⇄ H⁺(aq) and NO₂⁻(aq)

Step 2: Calculate the acid dissociation constant

Ka = [H⁺] × [NO₂⁻] / [HNO₂]

Ka = 0.022 × 0.022 / 0.68

Ka = 7.1 × 10⁻⁴

Step 3: Determine the strength of the acid

Since Ka is very small, nitrous acid is a weak acid, not completely dissociated in solution.

4 0
3 years ago
In need of help very soon please.
NemiM [27]

Answer:

hello, i hope this helps.

Explanation:

1 - group

2 - period

3 - periodic table

4 - family

5 - octet rule

6 - valence electrons

6 0
2 years ago
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