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Montano1993 [528]
3 years ago
10

Calculate the PH of a solution 0.030 MH2SO4

Chemistry
1 answer:
Zinaida [17]3 years ago
4 0

Answer:

pH= 2- log3

Explanation:

H2SO4 + H2O -> HSO4^(-) + H30^(+)

0.03M ___ ___

___ 0.03M 0.03M

H30^(+) : C = 0.03M

pH= - log( [H3O^(+)] ) => pH= - log {3× 10^(-2)} => pH = 2 - log3

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sveticcg [70]

Answer:

false the answer is woody

Explanation:

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3 years ago
Which of the following statements about matter is false?
4vir4ik [10]
The statement that is incorrect I believe would be A. The amount of matter increases slightly as water changes state.
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3 years ago
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The mass numbers for two isotopes are unequal because they have different numbers of
SSSSS [86.1K]
The mass numbers for two isotopes are unequal because they have different numbers of NEUTRONS.

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Calculate the masses of oxygen and nitrogen that are dissolved in of aqueous solution in equilibrium with air at 25 °C and 760 T
shtirl [24]

Explanation:

Let us assume that the volume of given aqueous solution is 7.5 L.

Therefore, according to Henry's law, the relation between concentration and pressure is as follows.

                C = \frac{P}{K_{h}}

where,   pressure (P) = 760 torr = 1 atm

According to Henry's law, constants for gases in water at 25^{o}C are as follows.

  p(O_{2}) = 0.21 atm = 0.21 bar

   p(N_{2}) = 0.78 atm = 0.78 bar

   K_{h} for O_{2} = 7.9 \times 10^{2} bar/mol

    K_{h} for N_{2} = 1.6 \times 10^{3} bar /mol

Since, 21% oxygen is present in air so, its mass will be 0.21 g. Similarly, 78% nitrogen means the mass of nitrogen is 0.78 g.

Therefore, concebtrations will be calculated as follows.

      C(O_{2}) = \frac{0.21}{7.9 \times 10^{2}} = 2.66 \times 10^-4 mol/L  

      C(N_2) = \frac{0.78}{1.6 \times 10^3} = 4.875 \times 10^-4 mol/L

Now, we will calculate the number of moles as follows.

         n(O_{2}) = 7.5 \times 2.66 \times 10^-4 = 1.995 \times 10^-3 mol

         n(N_2) = 7.5 \times 4.875 \times 10^-4 = 3.66 \times 10^-3 mol

As the molar mass of O_2 = 32 g/mol

Hence, mass of oxygen will be as follows.

         Mass of O_2 = 32 \times 1.995 \times 10^-3 \times 1000

                           = 63.84 mg

As the molar mass of N_{2} = 28

       Mass of N_{2} = 28 \times 3.66 \times 10^-3 = 102.5 mg

Thus, we can conclude that mass of oxygen is 63.84 mg  and nitrogen is 102.5 mg.

5 0
3 years ago
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From the combinations of substances listed below, which would most likely be miscible in
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Answer:

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