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solniwko [45]
4 years ago
8

In the following reaction, identify the oxidized species, reduced species, oxidizing agent, and reducing agent. Be sure to answe

r all parts. Cl2(aq) + 2 KI(aq) → 2 KCl(aq) +12(aq) Cl, is the (select) KI is the (select) and the (select) and the (select) A. A.
Chemistry
1 answer:
gtnhenbr [62]4 years ago
7 0

Answer :

Cl_2 is reduced species.

KI is oxidized species.

Cl_2 is oxidizing agent.

KI is reducing agent.

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Reducing agent : It is defined as the agent which helps the other substance to reduce and itself gets oxidized. Thus, it will undergo oxidation reaction.

Oxidizing agent : It is defined as the agent which helps the other substance to oxidize and itself gets reduced. Thus, it will undergo reduction reaction.

The balanced redox reaction is :

Cl_2(aq)+2KI(aq)\rightarrow 2KCl(aq)+I_2(aq)

The half oxidation-reduction reactions are:

Oxidation reaction : 2I^-\rightarrow I_2+2e^-

Reduction reaction : Cl_2^++2e^-\rightarrow 2Cl^-

From this we conclude that the 'KI' is the reducing agent that loses an electron to another chemical species in a redox chemical reaction and itself gets oxidized and 'Cl_2' is the oxidizing agent that gain an electron to another chemical species in a redox chemical reaction and itself gets reduced.

Thus, Cl_2 is reduced species.

KI is oxidized species.

Cl_2 is oxidizing agent.

KI is reducing agent.

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What is the pOH of 0.50 molar H3BO3?
Crazy boy [7]

<u>Answer:</u>

<em>A. 10.25</em>

<em></em>

<u>Explanation:</u>

Pkb =4.77

So pka = 14 - pka = 9.23

Ka =10^{-pka}

H_3 BO_3 (aq)+ H_2 O(l) H_2 BO_3^- (aq)+H_3 O^+ (aq)

Initial                0.50M                                 0                                 0

Change                  -x                                 +x                               +x

Equilibrium    0.50M-x                               +x                               +x

Ka =\frac {((x)(x))}{(0.50M-x)}

5.88\times10^{-10}= \frac {x^2}{(0.50M-x)}

(-x is neglected) so we get

5.88\times10^{-10}\times0.50=x^2\\\\x^2=2.94\times10^{-10}

x=\sqrt{x^2}=1.72\times10^{-5} M=H^3 O^{+}

pH=-log[H^3 O^+]\\\\pH=-log[1.72\times10^{-5}]\\\\pH=4.76

pOH = 14 - pH

= 14 - 4.76

pOH = 9.24 is the answer

Option A - 10.25 is the answer which is close to 9.24

4 0
3 years ago
2 H2S(g) ⇄ 2 H2(g) + S2(g) Kc = 9.3× 10-8 at 400ºC 0.47 moles of H2S are placed in a 3.0 L container and the system is allowed t
Anon25 [30]

<u>Answer:</u> The concentration of hydrogen gas at equilibrium is 1.648\times 10^{-3}M

<u>Explanation:</u>

We are given:

Initial moles of hydrogen sulfide gas = 0.47 moles

Volume of the container = 3.0 L

The molarity of solution is calculated by using the equation:

\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of the solution}}

So, \text{Initial molarity of hydrogen sulfide gas}=\frac{0.47}{3}=0.1567M

The given chemical equation follows:

                          2H_2S(g)\rightleftharpoons 2H_2(g)+S_2(g)

<u>Initial:</u>                  0.1567

<u>At eqllm:</u>           0.1567-2x       2x         x

The expression of K_c for above equation follows:

K_c=\frac{[H_2]^2[S_2]}{[H_2S]^2}

We are given:

K_c=9.3\times 10^{-8}

Putting values in above equation, we get:

9.3\times 10^{-8}=\frac{(2x)^2\times x}{(0.1567-2x)^2}\\\\x=8.24\times 10^{-4}

So, equilibrium concentration of hydrogen gas = 2x=(2\times 8.24\times 10^{-4})=1.648\times 10^{-3}M

Hence, the concentration of hydrogen gas at equilibrium is 1.648\times 10^{-3}M

7 0
3 years ago
6. The pOH of a solution of NaOH is 11.30. What is the [H+<br><br> ] for this solution?
sashaice [31]

Answer:

The [H⁺] for this soluton is 2*10⁻³ M

Explanation:

pH, short for Hydrogen Potential and pOH, or OH potential, are parameters used to measure the degree of acidity or alkalinity of substances.

The values ​​that compose them vary from 0 to 14 and the pH value can be directly related to that of pOH by means of:

pH + pOH= 14

In this case, pOH=11.30, so

pH + 11.30= 14

Solving:

pH= 14 - 11.30

pH= 2.7

Mathematically the pH is the negative logarithm of the molar concentration of the hydrogen or proton ions (H⁺) or hydronium ions (H₃O):

´pH= - log [H⁺] = -log [H₃O]

Being pH=2.7:

2.7= - log [H⁺]

[H⁺]= 10⁻² ⁷

[H⁺]=1.995*10⁻³ M≅ 2*10⁻³ M

<u><em>The [H⁺] for this soluton is 2*10⁻³ M</em></u>

8 0
4 years ago
The statement “the scientific process is open ended” means:
mina [271]

Answer:

think it helps you

<h2>Explanation:</h2>

<em><u>The statement “the scientific process is open ended” means: Would an element with 7 valence electrons be more or less reactive than an element with 3 valence electrons? Element 1 is a hard dark-red solid</u></em>

4 0
3 years ago
How many moles are in 48 molecules of H2O? What is the given amount?
Mkey [24]

Answer:

1.635

Explanation:

7 0
3 years ago
Read 2 more answers
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