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pochemuha
3 years ago
12

What are the relative ion concentrations in an acid solution?

Chemistry
1 answer:
Veronika [31]3 years ago
4 0
Answer is: more H⁺ ions than OH⁻ ions.
<span>An Arrhenius acid is a substance that dissociates in water to form hydrogen ions or protons. 
For example, hydrochloric acid dissociate in aqueous solution to form hydrogen ions (H</span>⁺) and chloride anion (Cl⁻):<span>
HCl(aq) → H</span>⁺(aq) + Cl⁻<span>(aq).
</span>

An Arrhenius base is a substance that dissociates in water to form hydroxide ions (OH⁻<span>). 
For example lithium hydroxide is an Arrhenius base:</span>

LiOH(aq) → Li⁺(aq) + OH⁻(aq).


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What is required for sexual reproduction to occur? an individual that produces sperm and an individual that produces eggs an ind
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Answer:

It is A.

Explanation:

I hope this helps have a great day.

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Which of the following is a correctly written chemical equation that demonstrates the conservation of mass?
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6 0
2 years ago
Suppose of barium acetate is dissolved in of a aqueous solution of ammonium sulfate. Calculate the final molarity of barium cati
ollegr [7]

Explanation:

Let us assume that the given data is as follows.

   mass of barium acetate = 2.19 g

   volume = 150 ml = 0.150 L    (as 1 L = 1000 ml)

   concentration of the aqueous solution = 0.10 M

Therefore, the reaction equation will be as follows.

        Ba(C_{2}H_{3}O_{2})_{2} \rightarrow Ba^{2+} + 2C_{2}H_{3}O^{-}_{2}

Hence, moles of C_{2}H_{3}O^{-}_{2} = 2 \times Ba(C_{2}H_{3}O_{2})_{2}  .......... (1)

As,   No. of moles = \frac{mass}{\text{molar mass}}

Hence, moles of Ba(C_{2}H_{3}O_{2})_{2} will be calculated as follows.                          

     No. of moles = \frac{mass}{\text{molar mass}}  

                          =  \frac{2.19 g}{255.415 g/mol}   (molar mass of Ba(C_{2}H_{3}O_{2})_{2} is 255.415 g/mol)            

                       = 8.57 \times 10^{-3}

    Moles of C_{2}H_{3}O^{-}_{2} = 2 \times 8.57 \times 10^{-3}

                          = 0.01715 mol

Hence, final molarity will be as follows.

              Molarity = \frac{\text{no. of moles}}{volume}

                             = \frac{0.01715 mol}{0.150 L}

                             = 0.114 M

Thus, we can conclude that final molarity of barium cation in the solution is 0.114 M.

5 0
3 years ago
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