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2 years ago

Question 10 of 25

Chemistry

2 answers:

kap26 [50]2 years ago

7 0

Answer:

Correct option would be A. Mg(s) → Mg2+ + 2e-

Explanation:

Marked as correct answer on Quiz ;) (A P E X)

Julli [10]2 years ago

7 0

Answer:

A. Mg(s) → Mg2+ + 2e

Explanation:

it was correct lol

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Why can liquid water never go over the boiling point temperature? Where is the energy (heat) that is being added going?

Afina-wow [57]

it's turned into steam.

It dissolves and turns into steam no explanation needed,it'sa simple answer,idk who reported it but i'll answer it again:)

7 0

3 years ago

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

Three of Louis Pasteur discovery

konstantin123 [22]

Pasteurization, vaccines against antrax and rabies, discovered the germ fermentation

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3 years ago

Most elements occur in nature as a single isotope. TRUE FALSE

cestrela7 [59]

Answer:

false

Explanation:

because elements never exist in nature as single isotopes they are always combined

8 0

3 years ago

Read 2 more answers

What mass of sodium bicarbonate do you start with

valentina_108 [34]

Answer:

84.007 g/mol

Explanation:

welcome

5 0

3 years ago