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alisha [4.7K]
3 years ago
10

Calculate the mass of ammonia (NH3) that contains a billion (1.00 * 10^12) hydrogen atoms.

Chemistry
1 answer:
Lynna [10]3 years ago
3 0

Answer:

The mass of ammonia (NH3) that contains 1.00 \times 10^12 hydrogen atoms is 9.4\times 10^{-12} g.

Explanation:

As 6.022\times 10^{23} atoms of hydrogen = 1 mole of the hydrogen atom

Therefore, 10^{12} atoms of hydrogen = \frac{1}{6.022 \times 10^{23}}\times 10^{12}=1.66\times 10^{-12} moles of the hydrogen atom.

Now, there are 3 moles of hydrogen atoms in 1 mole of ammonia (NH_3).

As the mass of 1 mole of ammonia is 17g, so

when there are 3 moles of hydrogen atoms, then the mass of ammonia = 17 g

Therefore, when there are 1.66\times 10^{-12} moles of hydrogen atoms, then the mass of ammonia = \frac{17}{3}\times 1.66\times 10^{-12}=9.4\times 10^{-12 g.

Hence, the mass of ammonia (NH_3) that contains 1.00 \times 10^12 hydrogen atoms is 9.4\times 10^{-12} g.

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In order to prepare 10 mL, 5 μM; <em> 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.</em>

In order to prepare 10 mL, 10 μM; <em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM; <em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM; <em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

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Using the dilution equation:

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<em>2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

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25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

<em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

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initial volume = 150/25 = 6 mL

<em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM from 25 μM stock,

25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

<em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

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