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bagirrra123 [75]
3 years ago
12

In the activated complex for a chemical reaction, what bonds are broken and what bonds are formed?

Chemistry
1 answer:
IceJOKER [234]3 years ago
8 0
Reactants bonds are broken and products bonds are formed
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Someone please help? I've had a go at this myself (see picture of working attached) but not sure where to go from there?
DiKsa [7]

Answer:

75 ml of 0.1M  base neutralizes 25 ml of 0.1M acid, which means the acid has 0.3 moles/L of H ion

but that means each molecule of the acid has 3 times as many H ions aH ions in a molecule of NaOH

which means the formula for the acid must be H3A and

the value of x in

HxA is 3

Explanation:

75ml of a solutipn of 0.1moL  l-1 NaOH neutralises 25ml of a solution of an acid. The formula of the acid is HxA and the concentration of the acid is 0.1mol l-1. What is the value of x?​

the concentration of both the and the base are the same at 0.1M

the base...NaOH has 0.1 moles/L of OH ion

75 ml of 0.1M  base neutralizes 25 ml of 0.1M acid, which means the acid has 0.3 moles/L of H ion

but that means each molecule of the acid has 3 times as many H ions aH ions in a molecule of NaOH

which means the formula for the acid must be H3A and the value of x in

HxA is 3

6 0
3 years ago
Question 14(multiple choice worth 2 points) a student had a sample of pure water, and added an unknown substance to it. the stud
pantera1 [17]
The answer is A its basic

5 0
3 years ago
Is topaz inorganic???????
cluponka [151]
<span>Inorganic Topaz is a silicate mineral of aluminium and fluorine. With the chemical formula Al2SiO4(F,OH)2.</span>
4 0
3 years ago
Acid rain can be destructive to both the natural environment and human-made structures. The equation below shows a reaction that
sleet_krkn [62]

Answer:

I would say the correct answer is <em><u>A 200.00 mol</u></em>

Explanation:

Hope this help

Have a nice day

8 0
3 years ago
Calculate the pH for each case in the titration of 50.0 mL of 0.230 M HClO ( aq ) 0.230 M HClO(aq) with 0.230 M KOH ( aq ) . 0.2
Vikentia [17]

Answer:

pH before addition of KOH = 4.03

pH after addition of 25 ml KOH = 7.40

pH after addition of 30 ml KOH = 7.57

pH after addition of 40 ml KOH = 8.00

pH after addition of 50 ml KOH = 10.22

pH after addition 0f 60 ml KOH = 12.3

Explanation:

pH of each case in the titration given below

(6) After addition of 60 ml KOH

Since addition of 10 ml extra KOH is added after netralisation point.

Concentration of solution after addition 60 ml KOH is calculated by

M₁V₁ = M₂V₂

or, 0.23 x 10 = (50 + 60)ml x M₂

or M₂ = 0.03 Molar

so, concentration of KOH = 0.03 molar

                               [OH⁻] = 0.03 molar

                                 pOH = 0.657

                                  pH = 14 - 0.657 = 13.34

5 0
3 years ago
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