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Ainat [17]
3 years ago
6

What are some ways a child could make money? (pls help this is part of school)

Chemistry
1 answer:
Damm [24]3 years ago
5 0

Answer:

you could get a job if you are of age if not ask your parents if you could do chores for allowance

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A weak monoprotic acid is titrated with 0.100 MNaOH. It requires 50.0 mL of the NaOH solution to reach the equivalence point. Af
larisa86 [58]

Explanation:

The given reaction is as follows.

        HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)

Hence, number of moles of NaOH are as follows.

        n = 0.05 L \times 0.1 M

           = 0.005 mol

After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.

             n = 0.025 L \times 0.1 M

                = 0.0025 mol

According to ICE table,

         HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)

Initial:     0.005 mol   0.0025 mol              0                  0

Change: -0.0025 mol  -0.0025 mol        +0.0025 mol

Equibm:   0.0025 mol    0                         0.0025 mol

Hence, concentrations of HA and NaA are calculated as follows.

          [HA] = \frac{0.0025 mol}{V}

        [NaA] = \frac{0.0025 mol}{V}

       [A^{-}] = [NaA] = \frac{0.0025 mol}{V}

Now, we will calculate the pK_{a} value as follows.

          pH = pK_{a} + log \frac{A^{-}}{HA}

       pK_{a} = pH - log \frac{[A^{-}]}{[HA]}

                  = 3.42 - log \frac{\frac{0.0025 mol}{V}}{\frac{0.0025}{V}}

                  = 3.42

Thus, we can conclude that pK_{a} of the weak acid is 3.42.

           

8 0
3 years ago
Please help!!
xxTIMURxx [149]

Answer:

In an experiment, a student transferred 4.50 mL of a liquid into a pre-weighed beaker (the weight of which was determined to be 35.986 g ).

Explanation:

<em>HOPE</em><em> </em><em>THIS</em><em> </em><em>HELPS</em><em> </em><em>YOU</em><em> </em>

<em>❤</em><em> </em><em>❤</em><em> </em><em>❤</em><em> </em><em>❤</em><em> </em><em>❤</em><em> </em><em>❤</em><em> </em><em>❤</em><em> </em><em>❤</em><em> </em>

6 0
3 years ago
How much energy (in joules) is released when 0.06 kilograms of mercury is condensed to a liquid at the same temperature? A. 51,3
expeople1 [14]
The amount of energy released when 0.06 kg of mercury condenses at the same temperature can be calculated using its latent heat of fusion which is the opposite of melting. Latent heat of fusion and melting can be used because they have the same magnitude, but opposite signs. Latent heat is the amount of energy required to change the state or phase of a substance. For latent heat, there is no temperature change. The equation is:

E = m(ΔH)

where:
m = mass of substance
ΔH = latent heat of fusion or melting

According to data, the ΔH of mercury is approximately 11.6 kJ/kg.

E = 0.06kg (11.6 kJ/kg) = 0.696 kJ or 696 J

The answer is D. 697.08 J. Note that small differences could be due to rounding off or different data sources.
3 0
3 years ago
Read 2 more answers
How do you write 561983 in scientific notation?
Artist 52 [7]

Answer:

5.61983 × 10^5

Explanation:

Move the decimal forward 5 spaces, each time doing so you add 10^(# of spaces moved, in this case 5)

7 0
3 years ago
What is the total charge of an iron nucleus
Dimas [21]
I believe it’s a positive charge
3 0
3 years ago
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