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Oksana_A [137]
3 years ago
13

Which chemical formula represents the hydroxide ion? H+ H3O+ H2O OH-

Chemistry
2 answers:
Elza [17]3 years ago
8 0
OH IS THE ANSWER to thia
Sophie [7]3 years ago
5 0
OH- is the hydroxide ion
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The price of tomatoes is 1.8 lb/$. How many $ would you need to buy 8.1 lb?​
sesenic [268]
If you divide 8.1 by 1.8 that leads to 4.5
7 0
3 years ago
The pH of a solution that is formed by the neutralization of 1.0M H2SO4(aq) and 1.0 M KOH(aq) is closest to
pshichka [43]

Answer:

b) 7

Explanation:

The pH of a solution produced by the neutralization reaction between 1M of H₂SO₄ and KOH with 1M is closest to 7.

pH is a standard for measuring the acidity and alkalinity of a solution. A solution that is acidic will have a pH less than 7, a neutral solution will have pH of 7 and a basic solution will have pH greater than 7.

What is a neutralization reaction?

  • It is an acid-base reaction in which hydrogen and hydroxide ions combines to form water.
  • Also a salt results from the combination of the other ions.

In this reaction a base simply neutralizes an acid and the solution becomes neutral before it goes into completion.

Therefore, a neutral solution will have pH of 7 or close to it.

3 0
3 years ago
The acid dissociation constant Ka of boric acid (H3BO3) is 5.8 times 10^-10. Calculate the pH of a 4.4 M solution of boric acid.
madam [21]

Answer: The pH of a 4.4 M solution of boric acid is 4.3

Explanation:

H_3BO_3\rightarrow H^+H_2BO_3^-

at t=0  cM              0             0

at eqm c-c\alpha        c\alpha          c\alpha  

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= 4.4 M and \alpha = ?

K_a=5.8\times 10^{-10}

Putting in the values we get:

5.8\times 10^{-10}=\frac{(4.4\times \alpha)^2}{(4.4-4.4\times \alpha)}

(\alpha)=0.000011

[H^+]=c\times \alpha

[H^+]=4.4\times 0.000011=4.8\times 10^{-5}M

Also pH=-log[H^+]

pH=-log[4.8\times 10^{-5}]=4.3

Thus pH of a 4.4 M H_3BO_3 solution is 4.3

3 0
3 years ago
The addition of 250.0 J to 30.0 g of copper initially at 22.0°C will change its temperature to what final value? (Specific heat
WINSTONCH [101]

Answer:

Final temperature = 43.53^{\circ} C

Explanation:

Given that,

Heat added, Q = 250 J

Mass, m = 30 g

Initial temperature, T₁ = 22°C

The Specific heat of Cu= 0.387 J/g °C

We know that, heat added due to the change in temperature is given by :

Q=mc\Delta T\\\\Q=mc(T_2-T_1)\\\\\dfrac{Q}{mc}=(T_2-T_1)\\\\T_2=\dfrac{Q}{mc}+T_1

Put all the values,

T_2=\dfrac{250}{30\times 0.387}+22\\\\=43.53^{\circ} C

So, the final temperature is equal to 43.53^{\circ} C.

8 0
3 years ago
Do simple molecules conduct electricity? and why?
ira [324]
Yes they have electrons
7 0
3 years ago
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