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Ahat [919]
2 years ago
13

Determine the normality of the following solutions note the species of interest is H 95 g of PO4 3- in 100mL solution

Chemistry
1 answer:
Aliun [14]2 years ago
5 0

Answer : The normality of the solution is, 30.006 N

Explanation :

Normality : It is defined as the number of gram equivalent of solute present in one liter of the solution.

Mathematical expression of normality is:

\text{Normality}=\frac{\text{Gram equivalent of solute}}{\text{Volume of solution in liter}}

or,

\text{Normality}=\frac{\text{Weight of solute}}{\text{Equivalent weight of solute}\times \text{Volume of solution in liter}}

First we have to calculate the equivalent weight of solute.

Molar mass of solute PO_4^{3-} = 94.97 g/mole

\text{Equivalent weight of solute}=\frac{\text{Molar mass of solute}}{\text{charge of the ion}}=\frac{94.97}{3}=31.66g.eq

Now we have to calculate the normality of solution.

\text{Normality}=\frac{95g}{31.66g.eq\times 0.1L}=30.006eq/L

Therefore, the normality of the solution is, 30.006 N

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Explanation:

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Hence, the order of steps:

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5 0
2 years ago
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Stabiliţi numerele de oxidare ale tuturor elementelor prezente în următoarele substanţe chimice, ţinând cont de principalele reg
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Answer:

a.

N = +2

O = -2

b.

Na = +1

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N = +5

c.

O = -2

Al = +3

P = +5

d.

O = -2

Ca = +2

C = +4

e

O = -2

Mn = +7

K = +1

f.

O = -2

K = +1

Cr = +6

g.

O = -2

Cl = +7

Cr = +1

h.

O = -2

H = +1

Cr = +5

i.

O = -2

H = +1

Cr = +3

k.

O = -2

H = +1

Cr = +1

Explanation:

A. NU

Numărul de oxidare a azotului = +2

Numărul de oxidare a oxigenului = -2

b. NaNO₃

Numărul de oxidare de Na = +1

Numărul de oxidare de O = -2

Numărul de oxidare de N = +5

c. AlPO₄

Numărul de oxidare de O = -2

Numărul de oxidare al Al = +3

Numărul de oxidare al P = +5

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Numărul de oxidare de O = -2

Numărul de oxidare de Ca = +2

Numărul de oxidare de C = +4

e. KMnO₄

Numărul de oxidare de O = -2

Numărul de oxidare al Mn = +7

Numărul de oxidare al lui K = +1

f. K₂Cr₂O₇

Numărul de oxidare de O = -2

Numărul de oxidare al lui K = +1

Numărul de oxidare al Cr = +6

g. HClO₄

Numărul de oxidare de O = -2

Numărul de oxidare al Cl = +7

Numărul de oxidare al Cr = +1

h. HClO₃

Numărul de oxidare de O = -2

Numărul de oxidare al lui H = +1

Numărul de oxidare al Cr = +5

i. HClO₂

Numărul de oxidare de O = -2

Numărul de oxidare al lui H = +1

Numărul de oxidare al Cr = +3

k. HClO

Numărul de oxidare de O = -2

Numărul de oxidare al lui H = +1

Numărul de oxidare al Cr = +1.

6 0
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Answer:

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Explanation:

7 0
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