Steroids are made up of what type of molecule?

1 answer:

6 0

Steroids composition:
They are made up of any of a large group of fat-soluble organic compounds, as the sterols, bile acids, and sex hormones, most of which have specific physiological action

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Calculate the percentage mass of hydrogen in hydrochloric acid, HCI

PIT_PIT [208]

Answer:

To find the mass percent of hydrogen in hydrogen chloride, we must divide the weight of the hydrogen atom alone by the weight of the entire molecule. Then we multiply by 100% to find the percentage. Thus, 2.77% of the mass of hydrogen chloride is hydrogen.

Explanation:

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2 years ago

When electricity is added to water, water breaks down and becomes hydrogen gas and oxygen gas. This is a chemical change because

LekaFEV [45]

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2 years ago

A natural water with a flow of 3800 m3/d is to be treated with an alum dose of 60 mg/L. Determine the chemical feed rate for the

svet-max [94.6K]

Explanation:

First, we will calculate the feed rate of alum as follows.

$\frac{\text{60 mg alum}}{\text{1 L water}} \times \frac{\text{1000 L water}}{1 m^{3}} \times \frac{3800 m^{3}}{day} \times \frac{\text{1 g alum}}{\text{1000 mg alum}}$

= 228000 g/day

Converting this amount into g/min as follows.

$\frac{228000 g}{1 day} \times \frac{1 day}{1440 min}$

= 158 g/min

Now, the chemical equation will be as follows.

$Al_{2}(SO_{4})_{3}.14H_{2}O \rightarrow 2Al(OH)_{3}(s) + 6H^{+} + 3SO^{2-}_{4} + 8H_{2}O$

$\frac{\text{30 mg alum}}{1 L} \times \frac{\text{1 mmol alum}}{\text{594 mg alum}} \times \frac{\text{3 mmol SO^{2-}_{4}}}{\text{1 mmol alum}}$

= 0.151 mmol $mmol SO^{2-}_{4}/L$

$\frac{0.151 mmol SO^{2-}_{4}}{L} \times \frac{\text{2 meq SO^{2-}_{4}}}{\text{1 mmol SO^{2-}_{4}}} \times \frac{\text{1 meq Alk}}{\text{1 meq SO^{2-}_{4}}} \times \frac{\text{50 mg CaCO_{3}}}{\text{1 meq Alk}}$

= 15.15 mg $CaCO_{3}$ /L

For precipitate:

$Al_{2}(SO_{4})_{3}.14H_{2}O \rightarrow 2Al(OH)_{3}(s) + 6H^{+} + 3SO^{2-}_{4} + 8H_{2}O$

$\frac{\text{30 mg alum}}{1 L} \times \frac{\text{1 mmol alum}}{\text{594 mg alum}} \times \frac{\text{2 mmol Al(OH)_{3}}}{\text{1 mmol alum}} \times \frac{\text{78 mg Al(OH)_{3}}}{\text{1 mmol Al(OH)_{3}}}$