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4 years ago

BRAINLIEST

Chemistry

1 answer:

Nat2105 [25]4 years ago

6 0

Answer:

[H₃O⁺] → 1×10¹¹

Second option

Explanation:

pH = - log [H₃O⁺]

So, 10^⁻pH = [H₃O⁺]

pH = 11

10⁻¹¹ = [H₃O⁺] → 1×10¹¹

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An unknown salt is either NaF, NaCl, or NaOCl. When 0.050 mol of the salt is dissolved in water to form 0.500 L of solution, the

victus00 [196]

Answer: The unknown salt is NaF

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of salt = 0.050 moles

Volume of solution = 0.500 L

Putting values in above equation, we get:

$\text{Molarity of salt}=\frac{0.050mol}{0.500L}\\\\\text{Molarity of salt}=0.1M$

To calculate the hydroxide ion concentration, we first calculate pOH of the solution, which is:

pH + pOH = 14

We are given:

pH = 8.08

$pOH=14-8.08=5.92$

To calculate pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

Putting values in above equation, we get:

$5.92=-\log[OH^-]$

$[OH^-]=10^{-5.92}=1.202\times 10^{-6}M$

The unknown salt given are formed by the combination of weak acid and strong acid which is NaOH

The chemical equation for the hydrolysis of $X^-$ ions follows:

$X^-(aq.)+H_2O(l)\rightleftharpoons HX(aq.)+OH^-(aq.);K_b$

Initial: 0.1

At eqllm: 0.1-x x x

Concentration of $OH^-=x=1.202\times 10^{-6}M$

The expression of $K_b$ for above equation follows:

$K_b=\frac{[OH^-][HX]}{[X^-]}$

Putting values in above expression, we get:

$K_b=\frac{(1.202\times 10^{-6})\times (1.202\times 10^{-6})}{(1-(1.202\times 10^{-6}))}\\\\K_b=1.445\times 10^{-11}M$

To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant

$K_b$ = Base dissociation constant = $1.445\times 10^{-11}$

Putting values in above equation, we get:

$10^{-14}=1.445\times 10^{-11}\times K_a\\\\K_a=\frac{10^{-14}}{1.445\times 10^{-11}}=6.92\times 10^{-4}$

We know that:

$K_a\text{ for HF}=6.8\times 10^{-6}$

$K_a\text{ for HCl}=1.3\times 10^{6}$

$K_a\text{ for HClO}=3.0\times 10^{-8}$

So, the calculated $K_a$ is approximately equal to the $K_a$ of HF

Hence, the unknown salt is NaF

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3 years ago

Calculate the energy (in kJ) required to heat 10.1 g of liquid water from 55 oC to 100 oC and change it to steam at 100 oC. The

Maksim231197 [3]

Answer:

$\large\boxed{\large\boxed{24.6kJ}}$

Explanation:

1. Energy to heat the liquid water from 55ºC to 100ºC

$Q=m\times C\times \Delta T$

m = 10.1g

C = 4.18g/JºC

ΔT = 100ºC - 55ºC = 45ºC

$Q=10.1g\times 4.18J/g\ºC\times 45\ºC=1,899.81J$

2. Energy to change the liquid to steam at 100ºC

$L=\lambda \times n$

λ = 40.6kJ/mol

n = 10.1g / 18.015g/mol = 0.5606mol

$L=40.6kJ/mol\times 0.5604mol=22.76214kJ=22,762.14J$

3. Total energy

$1,899.81J+22,762.14J=24,661.95J\approx24,662J\approx24.6kJ$

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The answer is the moon

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Answer:

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Explanation:

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Answer:

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