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aleksklad [387]
3 years ago
13

If nitrogen (N) has 2 naturally occurring isotopes, nitrogen-14 (78.3%) and nitrogen-16 (21.7%), what is its average r.a.m.?

Chemistry
1 answer:
leva [86]3 years ago
5 0

Answer:

14.434 r.a.m.

Explanation:

  • The atomic mass of an element is a weighted average of its isotopes in which the sum of the abundance of each isotope is equal to 1 or 100%.

∵ The atomic mass of N = ∑(atomic mass of each isotope)(its abundance)

∴ The atomic mass of N = (atomic mass of N-14)(abundance of N-14) + (atomic mass of N-16)(abundance of N-16)

atomic mass of N-14 = 14.0 r.a.m, abundance of N-14 = percent of N-14/100 = 78.3/100 = 0.783.

atomic mass of N-16 = 16.0 r.a.m, abundance of N-16 = percent of N-16/100 = 21.7/100 = 0.217.

∴ The atomic mass of N = (atomic mass of N-14)(abundance of N-14) + (atomic mass of N-16)(abundance of N-16) = (14.0 r.a.m)(0.783) + (16.0 r.a.m)(0.217) = 14.434 r.a.m.

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Strike441 [17]

Answer:

Fe_6O_9

Explanation:

Hello there!

In this case, since these problems about formulas, firstly require the determination of the empirical formula, assuming that the given percentages are masses, we can calculate the moles and mole ratio of oxygen to iron as shown below:

n_{Fe}=70/55.85=1.25\\\\n_O=30/16=1.875

In such a way, by rounding to the first whole number we multiply by 8 and divide by 5 to obtain:

Fe_{2}O_{3}

Whose molar mass is 159.69 g/mol and the mole ratio of the molecular to the empirical formula is:

479.1/159.69=3

Therefore, the molecular formulais:

Fe_6O_9

Regards!

4 0
2 years ago
A solution used to chlorinate a home swimming pool contains 7% chlorine by mass. An ideal chlorine level for the pool is one par
deff fn [24]

Answer:

Volume of chlorine = 61.943 mL

Explanation:

Given:

Volume of the water in the Pool = 18,000 gal

also,

1 gal = 3785.412 mL

thus,

Volume of water in pool = 18,000 × 3785.412 = 68,137,470 mL

Density of water = 1.00 g/mL

Therefore,

The mass of water in the pool = Volume × Density

or

The mass of water in the pool = 68,137,470 mL × 1.00 g/mL = 68,137,470 g

in terms of million = \frac{\textup{68,137,470}}{\textup{ 1,000,000}}

or

= 68.13747 g

also,

1 g of chlorine is present per million grams of water

thus,

chlorine present is 68.13747 g

Now,

volume = \frac{\textup{Mass}}{\textup{Density}}

or

Volume of chlorine = \frac{\textup{68.13747 g}}{\textup{1.10 g/mL}}

or

Volume of chlorine = 61.943 mL

3 0
3 years ago
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Flura [38]
Potential energy......
5 0
3 years ago
Identify the oxidizing and reducing agent in the following reaction, and determine which element is oxidized and which is reduce
erma4kov [3.2K]

Answer:

Explanation:

Fe⁺²(aq) + ClO₂(aq) → Fe⁺³(aq) + ClO₂⁻(aq)

Here oxidation number of Fe is increased from +2 to +3 , so Fe is oxidised .

The oxidation number of Cl is reduced from + 4 to +3  so Cl is reduced .

So ClO₂(aq) is oxidising agent and Fe⁺²(aq) is reducing agent .

8 0
3 years ago
Please anwser this question will me thanked
Nataly_w [17]
The answer should be "by convection" not by radiation. 
7 0
3 years ago
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