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Maru [420]
3 years ago
6

How much 6.0 M HNO3 is needed to neutralize 39 mL of 2.0 M KOH?

Chemistry
2 answers:
Eva8 [605]3 years ago
7 0
<span>KOH+HNO3--->KNO3+H2O
CA=6.0M, CB=2.0M, VA=?, VB=39ml, na=1, nb=1
CAVA/CBVB=na/nb
6*VA/2*39=1/1
6VA/78=1
6VA=78
VA=78/6. VA=13ml.
VA=13ml. </span>
san4es73 [151]3 years ago
4 0

Answer : The volume of HNO_3 needed are, 0.013 L

Explanation :

According to the neutralization law,

M_1V_1=M_2V_2

where,

M_1 = molarity of HNO_3 = 6.0 M = 6.0 mole/L

M_2 = molarity of KOH = 2.0 M = 2.0 mole/L

V_1 = volume of HNO_3 = ?

V_2 = volume of KOH = 39 ml = 0.039 L

conversion : 1 L = 1000 ml

Now put all the given values in the above formula, we get the volume of HNO_3

6.0mole/L\times V_1=2.0mole/L\times 0.039L

V_1=0.013L

Therefore, the volume of HNO_3 needed are, 0.013 L

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Answer:

When balancing redox reactions under basic conditions in aqueous solution, the first step is to balance oxygen.

Explanation:

Oxidation-reduction reactions or redox reactions are those in which an electron transfer occurs between the reagents. An electron transfer implies that there is a change in the number of oxidation between the reagents and the products.

The gain of electrons is called reduction and the loss of electrons oxidation. That is to say, there is oxidation whenever an atom or group of atoms loses electrons (or increases its positive charges) and in the reduction an atom or group of atoms gains electrons, increasing its negative charges or decreasing the positive ones.

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