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Nataly [62]
3 years ago
9

Magnesium has three stable isotopes. One has 12 neutrons the second has 13 and the third has 14. Which one is the most abundant?

How do you know?
Chemistry
1 answer:
fiasKO [112]3 years ago
7 0

Answer:

12

Explanation:

If you look at the periodic table, you can see that magnesium has an atomic mass of 24.305.  Subtract the number of protons from this to get neutrons.

24.305 - 12 = 12.305

12.305 is the average amount of neutrons a magnesium atom can have.  A magnesium atom can have 12, 13, or 14 neutrons.  The average, based on the natural abundance, is 12.305.  To find which isotope is more abundant, you need to pick the number that is closest to the average.  This would make the most abundant isotope the one with 12 neutrons.

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Rainfall is an abiotic factor. If rainfall in an ecosystem decreases, what happens to the amount of vegetation?
natali 33 [55]

Answer:

B. The amount of vegetation decreases.

Explanation:

plants can't grow without rain

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Select the classification for the following reaction. 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) a. precipitation b. acid-base c. redo
Alexeev081 [22]

Answer:

The answer to your question: I think is letter c

Explanation:

                       2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

a.- Precipitation: is incorrect because for this kind of reactions one of the products  must be a precipitate and in the reaction we see that NAOH is soluble in water (aq) and H2 is a gas, none of them precipitate.

b. acid - base: the reactants must be an acid and a base, and the reactants of this reaction has a base but it's in the products.

c. Redox: This reaction is a redox reaction because for this reaction the oxidation numbers of the reactants must change and this happen in this reaction (H changes from +1 to 0) and Na changes from 0 to +1.

d. Combination. In this reaction the reactants combine to form a new compound and this doesn't happen here.

e. None of the above. is false because letter C is correct.

8 0
3 years ago
Money management becomes more important when you are responsible for paying all your own expenses.
damaskus [11]
Hi there!

You're answer is going to be -

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5 0
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How many lone pairs are on the central atom in GeH4?
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7 0
3 years ago
1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium
lilavasa [31]

Answer:

This question is incomplete, here's the complete question:

1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium ion present in 20.0 mL of each buffer.

Buffer A

Mass of sodium acetate used: 0.3730 g

Actual ph of the buffer 5.27

volume of the buffer used in buffer capacity titration 20.0 mL

Concentration of standardized NaOH 0.100M

moles of Naoh needed to change the ph by 1 unit for the buffer 0.00095mol

the buffer capacity 0.0475 M

Buffer B

Mass of sodium acetate used 1.12 g

Actual pH of the buffer 5.34

Volume of the buffer used in buffer capacity titration 20.0 mL

Concentration if standardized NaOH 0.100 M

moles of Naoh needed to change the ph by 1 unit 0.0019 mol

the buffer capacity 0.095 M

2.) A change of pH by 1 unit means a change in hydronium ion concentration by a factor of 10. Calculate the number of moles of NaOH that would theoretically be needed to decrease the moles of hydronium you calculated in #1 by a factor of 10 for each buffer. Are there any differences between your experimental results and the theoretical calculation?

3.) which buffer had a higher buffer capacity? Why?

Explanation:

Formula,

moles = grams/molar mass

molarity = moles/L of solution

1. Buffer A

molarity of NaC2H3O2 = 0.3731 g/82.03 g/mol x 0.02 L = 0.23 M

molarity of HC2H3O2 = 0. 1 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.23/0.1)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

Buffer B

molarity of NaC2H3O2 = 1.12 g/82.03 g/mol x 0.02 L = 0.68 M

molarity of HC2H3O2 = 0.3 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.68/0.3)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

2. let x moles of NaOH is added,

Buffer A,

pH = 5.10

[H3O+] = 7.91 x 10^-6 M

new pH = 4.10

new [H3O+] = 7.91 x 10^-5 M

moles of NaOH to be added = (7.91 x 10^-5 - 7.91 x 10^-6) x 0.02 L

= 1.42 x 10^-6 mol

3. Buffer B with greater concentration of NaC2H3O2 and HC2H3O2 has higher buffer capacity as it resists pH change to a wider range due to addition of acid or base to the system as compared to low concentration of Buffer A

5 0
3 years ago
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