NASA uses radar to track as much space junk as possible, but it's estimated that millions of pieces

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]

<u>Answer:</u> The unknown salt is NaF

<u>Explanation:</u>

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:

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

<u>Initial:</u> 0.1

<u>At eqllm:</u> 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

6 0

3 years ago

How many atoms of fluorine are present in a molecule of carbon tetrafluoride CF4

Daniel [21]

There are 4, 1 Carbon 4 Fluorine

3 0

3 years ago

The oxidation state of phosphorus is +3 in

IgorC [24]

Answer:

b) Phosphorus acid

Explanation:

To distinguish the type of acid of phosphorus with the oxidation state of +3, we need to be familiar with the chemical formula of each of the compounds:

Orthophosphoric acid H₃PO₄

Phosphorus acid H₃PO₃

Metaphosphoric acid HPO₃

Phyrophosphoric acid H₄P₂O₇

Now that we know the formula of the given compounds, the algebraic sum of all the oxidation numbers of all atoms in a neutral compound is zero:

Only phosphorus acid yielded an oxidation state of +3 for phosphorus in the compound.

H₃PO₃:

we know the oxidation state of H = +1

O = -2

The oxidation state of P is unknown. We can express this as an equation:

3(+1) + P + 3(-2) = 0

3 + P -6 = 0

P-3 = 0

P = +3

6 0

3 years ago

A sample of solid sodium hydroxide, weighing 13.20 grams is dissolved in deionized water to make a solution. What volume in mL o

Andreas93 [3]

<h3>Answer:</h3>

2.809 L of H₂SO₄

<h3>Explanation:</h3>

Concept tested: Moles and Molarity

In this case we are give;

Mass of solid sodium hydroxide as 13.20 g

Molarity of H₂SO₄ as 0.235 M

We are required to determine the volume of H₂SO₄ required

<h3>First: We need to write the balanced equation for the reaction.</h3>

The reaction between NaOH and H₂SO₄ is a neutralization reaction.
The balanced equation for the reaction is;

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

<h3>Second: We calculate the umber of moles of NaOH used </h3>

Number of moles = Mass ÷ Molar mass
Molar mass of NaOH is 40.0 g/mol

Therefore;

Moles of NaOH = 13.20 g ÷ 40.0 g/mol

= 0.33 moles

<h3>Third: Determine the number of moles of the acid, H₂SO₄</h3>

From the equation, 2 moles of NaOH reacts with 1 mole of H₂SO₄
Therefore, the mole ratio of NaOH: H₂SO₄ is 2 : 1.
Thus, Moles of H₂SO₄ = moles of NaOH × 2

= 0.33 moles × 2

= 0.66 moles of H₂SO₄

<h3>Fourth: Determine the Volume of the acid, H₂SO₄ used</h3>

When given the molarity of an acid and the number of moles we can calculate the volume of the acid.

That is; Volume = Number of moles ÷ Molarity

In this case;

Volume of the acid = 0.66 moles ÷ 0.235 M

= 2.809 L

Therefore, the volume of the acid required to neutralize the base,NaOH is 2.809 L.

7 0

4 years ago

Which of these substances has the lowest pH? 0.5 M HBr, pOH = 13.5 0.05 M HCl, pOH = 12.7 0.005 M KOH, pOH = 2.3

vagabundo [1.1K]

<h3><u>Answer;</u></h3>

0.5 M HBr, pOH = 13.5 ; Has the lowest pH

<h3><u>Explanation;</u></h3>

From the question;

pH = -Log [OH]

or pH = 14 - pOH

Therefore;

For 0.5 M HBr

[H+] = 0.5 M

pH = - Log [0.5]

= 0.30

For; pOH = 13.5

pH = 14 - pOH

= 14 -13.5

= 0.5

For; 0.05 M HCl

pH = - log [H+]

[H+] = 0.05

pH = - Log [0.05]

= 1.30

For; pOH = 12.7

pH = 14 -pOH

= 14 -12.7

= 1.30

For; 0.005 M KOH,

pOH = - log [OH]

[OH-] = 0.005

pOH = - Log 0.005

= 2.30

pH = 14 - 2.30

= 11.7

For; pOH = 2.3

pH = 14 -pOH

= 14- 2.3

= 11.7

6 0

3 years ago

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