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Gnom [1K]
2 years ago
10

A chemist is using a solution of hno3 that has a ph of 3.75. what is [h ] for the solution?

Chemistry
1 answer:
kap26 [50]2 years ago
4 0

A chemist is using a solution of HNO₃ that has a pH of 3.75. what is [H⁺] for the solution is 1.7 × 10⁻⁴M.

<h3>How do we calculate the [H⁺]?</h3>

Concentration of H⁺ ion will be calculated by using the below equation of pH as:

pH = -log[H⁺]

or [H⁺] = 10^{-pH}

Given that, pH = 3.75

So concentration of H⁺ ion will be calculated as:

[H⁺] = 10^{-3.75}

[H⁺] = 1.7 × 10⁻⁴M

Hence concentration of H⁺ ion is 1.7 × 10⁻⁴M.

To know more about pH & [H⁺], visit the below link:
brainly.com/question/8758541

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Which compound could serve as a reactant in a neutralization reaction ?
Veseljchak [2.6K]
 The   compound that  could serve  as  a reactant   in the neutralization reaction is    H2SO4
     
  Explanation
   Neutralization reaction occur  between an acids  and a base. H2SO4             (  sulfuric acid)  is  a strong  acid.  It can be neutralized by  strong base such as NaOH ( sodium  hydroxide)

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3 years ago
May someone pls help me with this!
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4 0
3 years ago
A: -3.45<br> B: 17.45 <br> C: 10.55<br> D: 3.55
Minchanka [31]

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6 0
4 years ago
Three positive charges lie on the x axis: q1 = 1 × 10−8 C at x1 = 1 cm, q2 = 2 × 10−8 C at x2 = 2 cm, and q3 = 3 × 10−8 C at x3
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Answer: Option (4) is the correct answer.

Explanation:

Relation between potential energy and charge is as follows.

                 U = \frac{1}{4 \pi \epsilon_{o}}[\frac{q_{1}q_{2}}{r_{12}} + \frac{q_{2}q_{3}}{r_{23}} + \frac{q_{3}q_{1}}{r_{31}}]

As it is given that q_{1} = 1 \times 10^{-8} C, q_{2} = 2 \times 10^{-8} C, and q_{3} = 3 \times 10^{-8} C.

        Distance between the charges = 1 cm = 1 \times 10^{-2} m  (as 1 cm = 0.01 m)

Hence, putting these given values into the above formula as follows.

                 U = \frac{1}{4 \pi \epsilon_{o}}[\frac{q_{1}q_{2}}{r_{12}} + \frac{q_{2}q_{3}}{r_{23}} + \frac{q_{3}q_{1}}{r_{31}}]

            = 9 \times 10^{9} [\frac{1 \times 10^{-8} \times 2 \times 10^{-8}}{10^{-2}} + \frac{2 \times 10^{-8} \times 3 \times 10^{-8}}{10^{-2}} + \frac{3 \times 10^{-8} \times 1 \times 10^{-8}}{10^{-2}}]    

            = 9 \times 10^{9} [2 + 6 + 1.5]

            = 85.5 \times 10^{-5} J

            = 0.00085 J

Thus, we can conclude that the potential energy of this arrangement, relative to the potential energy for infinite separation, is about 0.00085 J.              

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3 years ago
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