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aliya0001 [1]
3 years ago
11

How many moles of solute are present in 150mL of 0.30 mol/L NaOH solution ourse

Chemistry
1 answer:
Y_Kistochka [10]3 years ago
6 0

Answer:

0.045 moles of NaOH

Explanation:

When this units are together mol/L, we talk about M, molarity.

This kind of concentration shows the relation between moles of solute which are contained in 1L of solution.

In this case, our NaOH solution is 0.30 M

M = mol/L

Then, M . L = mol

We convert volume to L → 150 mL . 1L /1000 mL = 0.150L

0.150 L . 0.30mol/L = 0.045 moles

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Answer:

Man-made resources

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

Cl_2 is reduced species.

KI is oxidized species.

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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.

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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.

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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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Paladinen [302]
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Since you are asked the amount of atom, you need to multiply the unit in mole with Avogadro number. The calculation should be: 0.74 * 6.02 * 10^23= 4.45* 10^23 molecules</span>
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