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bixtya [17]
3 years ago
15

What is the atomic number and the mass number of an atom with 11 protons and 12 neutrons?

Chemistry
1 answer:
sleet_krkn [62]3 years ago
8 0

Answer:

e

Explanation:

e

e

e

e

ee

e

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Anthony's homework assignment is to demonstrate that an orange has already undergone a chemical change. Which of the following
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B. Rotten orange is the correct answer. Hope this helps!
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4 years ago
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Which statement is true for a solution when its concentration of hydroxide ions increases? It becomes more acidic. It becomes le
makvit [3.9K]

Answer:

Its pH value increases.

Explanation:

pH is the measure of alkalinity or acidity of a compound.

pH = - log [H+]

and pH + pOH = 14

where pOH is the measure of basicity of a solution, given by -log[OH-]

As a solution gets more basic that is higher [OH-], the pH increases, and on the other hand, as the pH of a solution decreases by one pH unit, the concentration of H+ increases by ten times.

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Which change will always take place during nuclear fission?
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Answer:

um pretty sure this is it

Explanation:

The fission process also releases extra neutrons, which can then split additional atoms, resulting in a chain reaction that releases a lot of energy.

5 0
2 years ago
Assume the pH of household bleach is 12. What is the hydronium ion concentration in household bleach? What is it hydroxide ion c
Ostrovityanka [42]

Answer:

[H3O+] = 1.0*10^-12 M

[OH-] = 0.01 M

Explanation:

We can use the following equation to find the hydronium ion concentration.  Plug in the pH and solve for H3O+.

pH = -log[H3O+]

<u>[H3O+] = 1.0*10^-12 M</u>

Now, to find the hydroxide ion concentration we will use the two following equations.

14 = pH + pOH

pOH = -log[OH-]

14 = 12 + pOH

pOH = 2

2 = -log[OH-]

<u>[OH-] = 0.01 M</u>

8 0
3 years ago
A chemist makes 940. mL of sodium carbonate (Na_2CO_3) working solution by adding distilled water to 200. mL of a 0.171 m stock
Alexxandr [17]

Answer:

The answer is 0.0698 M

Explanation:

The concentration was prepared by a serial dilution method.

The formula for the preparation I M1V1 = M2V2

M1= the concentration of the stock solution = 0.171 M

V1= volume of the stock solution taken = 200 mL

M2 = the concentration produced

V2 = the volume of the solution produced = 940 mL

Substitute these values in the formula

0.171 × 200 = 490 × M2

34.2 = 490 × M2

Make M2 the subject of the formula

M2 = 34.2/490

M2 = 0.069795

M2 = 0.0698 M ( 3 s.f)

The concentration of the Chemist's working solution to 3 significant figures is 0.0698M

6 0
3 years ago
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