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Lorico [155]
2 years ago
7

A solution of SO2 in water contains 0.00023 g of SO2 per liter of solution. What is the concentration of SO2 in ppm? in ppb?

Chemistry
2 answers:
antoniya [11.8K]2 years ago
8 0

Answer:

0.23 ppm , 230 ppb  

Explanation:

1 ppm (parts per million) is equivalent to 1 mg of SO₂ in 1 L of the water

Given mass of SO₂ = 0.00023 g = 0.00023 x 1000 mg (1g = 1000 mg) = 0.23 mg

So 0.23 mg/ L = 0.23 ppm

1 ppb (parts per billion) (10⁹) = 1000 ppm (parts per million) (10⁶)

0.23 ppm x 1000 = 230 ppb  

Mnenie [13.5K]2 years ago
5 0

Answer:

= 230 ppb

Explanation:

Considering that;

1ppm = 1mg/L  

Then;

0.00023g = 0.23mg  

Therefore;

0.00023 g/L = 0.23 mg/L

0.23 mg/L = 0.23 ppm

1 ppm = 100 ppb

Therefore;

0.23 ppm = 0.23 ×1000

                = 230 ppb

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TiliK225 [7]

Answer:

The empirical formula of this substance is:

C_3H_6O_2

Explanation:

To find the empirical formula of this substance we need the molecular weight of the elements Carbon, Hydrogen and Oxygen, we can find this information in the periodic table:

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- H: 1.00 g/mol

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With the information in this exercise we can suppose in 100 g of the substance we have:

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H: 8.16 g

O: 43.2 g (100 g - 48.64g - 8.16g= 43.2 g)

Now, we need to divide these grams by the molecular weight:

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We need to divide these results by the minor result, in this case O=2.70 mol

C=\frac{4.05mol}{2.70mol}= 1.5\\H= \frac{8.16mol}{2.70 mol} = 3.02 \\O=\frac{2.70mol}{2.70mol} = 1

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C= 1.5*2=3\\H= 3.02*2= 6.04 \\O= 1*2=2

This numbers are very close to integer numbers, so we can find the empirical formula as  subscripts in the chemical formula:

C_3H_6O_2

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