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xeze [42]
3 years ago
8

Carbon dioxide in the atmosphere dissolves in raindrops to produce carbonic acid (H2CO3), causing the pH of clean, unpolluted ra

in to range from about 5.2 to 5.6.
What is the range of [H+] in the raindrops?
Chemistry
1 answer:
anastassius [24]3 years ago
6 0

Answer:

The range of [H⁺] is from 2.51 x 10⁻⁶ M to 6.31 x 10⁻⁶ M,

Explanation:

To answer this problem we need to keep in mind the <u>definition of pH</u>:

  • pH = -log [H⁺]

So now we <u>calculate [H⁺] using a pH value of 5.2 and of 5.6</u>:

  • 5.2 = -log [H⁺]

-5.2 = log [H⁺]

10^{-5.2} = [H⁺]

6.31 x 10⁻⁶ M = [H⁺]

  • 5.6 = -log [H⁺]

-5.6 = log [H⁺]

10^{-5.6} = [H⁺]

2.51 x 10⁻⁶ M = [H⁺]

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Natural gas (CH4) has a molar mass of 16.0 g/mole. You started out the day with a tank containing 200.0 g of natural gas. At the
hodyreva [135]

Considering the definition of molar mass, the moles of gas used are 10.625 moles.

<h3>Definition of molar mass</h3>

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

<h3>Amount of moles used</h3>

Natural gas has a molar mass of 16.0 g/mole.

You started out the day with a tank containing 200.0 g of natural gas.  So, you can apply the following rule of three: If by definition of molar mass 16 grams are contained in 1 mole, 200 grams are contained in how many moles?

amount of moles at the beginning=\frac{200 gramsx1 mole}{16 grams}

<u><em>amount of moles at the beginning= 12.5 moles</em></u>

At the end of the day, your tank contains 30.0 g of natural gas. So, you can apply the following rule of three: If by definition of molar mass 16 grams are contained in 1 mole, 30 grams are contained in how many moles?

amount of moles at the end=\frac{30 gramsx1 mole}{16 grams}

<u><em>amount of moles at the end= 1.875 moles</em></u>

The number of moles used will be the difference between the number of moles used initially and the contents at the end of the day.

moles used= amount of moles at the beginning - amount of moles at the end

moles used= 12.5 moles - 1.875 moles

<u><em>moles used= 10.625 moles</em></u>

<u><em /></u>

Finally, the moles of gas used are 10.625 moles.

Learn more about molar mass:

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