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2 years ago

In which compound does chlorine have the highest oxidation number?(1) NaClO (3) NaClO3(2) NaClO2 (4) NaClO4

Chemistry

1 answer:

lana [24]2 years ago

3 0

Answer:

NaClO₄

Solution:

Let us calculate the oxidation state of Chlorine in each compound one by one;

1) NaClO

O.N of Na = +1

O.N of O = -2

So,

(+1) + (Cl) + (-2) = 0

1 + Cl - 2 = 0

Cl = -1 + 2

Cl = +1

2) NaClO₃

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

(+1) + (Cl) - 6 = 0

1 + Cl - 6 = 0

Cl = -1 + 6

Cl = +5

3) NaClO₂

(+1) + (Cl) + (-2)2 = 0

(+1) + (Cl) + (-4) = 0

1 + Cl - 4 = 0

Cl = -1 + 4

Cl = +3

4) NaClO₄

(+1) + (Cl) + (-2)4 = 0

(+1) + (Cl) + (-8) = 0

1 + Cl - 8 = 0

Cl = -1 + 8

Cl = +7

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IF(subscript 5)

Explanation:

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1 year ago

All of the following reactions can be described as displacement reactions except:____________.

Lady_Fox [76]

Answer:

Explanation:

The reaction that is not a displacement reaction from all the options is $C_6H_6_{(l)} + Cl_{2(g)} --> C_6H_5Cl_{(l)} + HCl_{(g)}$

In a displacement reaction, a part of one of the reactants is replaced by another reactant. In single displacement reactions, one of the reactants completely displaces and replaces part of another reactant. In double displacement reaction, cations and anions in the reactants switch partners to form products.

Options a, c, d, and e involves the displacement of a part of one of the reactants by another reactant while option b does not.

Correct option = b.

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2 years ago

A solution has a hydroxide-ion concentration of 0.0040 M. What is the pOH of the solution? A solution has a pH value of 3.66. Wh

hodyreva [135]

Answer:

1. pOH = 2.4

2. pOH = 10.34

3. pH = 2.14

Explanation:

1. Determination of the pOH.

Concentration of Hydroxide ion, [OH-] = 0.004 M

pOH =?

pOH = - log [OH-]

pOH = - log (0.004)

pOH = 2.4

Therefore, the pOH of the solution is 2.4

2. Determination of the pOH.

pH = 3.66

pOH =?

pH and pOH are related by the following equation:

pH + pOH = 14

With the above formula, we can obtain the pOH of the solution as follow:

pH = 3.66

pOH =?

pH + pOH = 14

3.66 + pOH = 14

Collect like terms

pOH = 14 - 3.66

pOH = 10.34

Therefore, the pOH of the solution is 10.34

3. Determination of the pH.

Molarity of HCl = 0.0072 M

Concentration of Hydrogen ion, [H+] =?

Thus, we can obtain the concentration of Hydrogen ion, [H+] as follow:

HCl(aq) —> H+(aq) + Cl-(aq)

From the balanced equation above,

1 mole of HCl produced 1 mole of H+.

Therefore, 0.0072 M HCl will also produce 0.0072 M H+.

Therefore, the concentration of Hydrogen ion, [H+] in the solution is 0.0072 M.

Finally, we shall determine the pH of the solution as follow:

Concentration of Hydrogen ion, [H+] = 0.0072 M.

pH =?

pH = - log [H+]

pH = - log (0.0072)

pH = 2.14

Therefore, the pH of the solution is 2.14

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2 years ago

During an experiment water was heated in a container covered with a glass lid. The following observations were recorded.

Hitman42 [59]

It is a physical change because it only changed states

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2 years ago

Read 2 more answers

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

6 0

2 years ago