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Damm [24]
3 years ago
14

This method of treating water kills microbes and removes other contaminates such as heavy metals.

Chemistry
1 answer:
Ivenika [448]3 years ago
3 0
The answer is (B).
Hope this helps :).
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How many moles of sulfur dioxide (SO2) are required to produce 5.0 moles of sulfur (S) according to the following balanced equat
scoundrel [369]

Answer:

1.67 moles

Explanation:

From the balanced equation of reaction:

                     SO_2 + 2H_2S -> 3S + 2H_2O

1 mole of sulfur dioxide, SO2, is required to produce 3 moles of sulfur, S.

<em>If 1 mole SO2 = 3 moles S, then, how many moles of SO2 would be required for 5 moles S?</em>

     Moles of SO2 needed = 5 x 1/3

                   = 5/3 or 1.67 moles

Hence, <u>1.67 moles of SO2 would be required to produce 5.0 moles of S.</u>

3 0
2 years ago
Read 2 more answers
What is the IUPAC name of the following compounds i. CO₂ ii. NO₂ iii. HNO3 iv. H₂SO4 v. K-Cr₂O vi. [Fe(CN)
Zielflug [23.3K]

Explanation:

I) carbon dioxide

ii) nitrogen dioxide

iii) nitric acid

iv) sulphric acid

v) Potassium dichromate

vi) hexacyanoferrate (III) ion( not sure)

8 0
2 years ago
Normal body temperature is 98.6 degrees fahrenheit is what degrees Celsius
anyanavicka [17]

Answer: 37

Explanation:

8 0
3 years ago
Read 2 more answers
Jim and Jan were racing in a 400 meter event. Jim ran the race in 44.32 seconds. Jan ran the race in 42.21 seconds. Calculate th
nordsb [41]

Answer:

2.11

Explanation:

44.32- 42.21=2.11

Hope this helps

4 0
2 years ago
Read 2 more answers
The acid dissociation constant Ka of boric acid (H3BO3) is 5.8 times 10^-10. Calculate the pH of a 4.4 M solution of boric acid.
madam [21]

Answer: The pH of a 4.4 M solution of boric acid is 4.3

Explanation:

H_3BO_3\rightarrow H^+H_2BO_3^-

at t=0  cM              0             0

at eqm c-c\alpha        c\alpha          c\alpha  

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= 4.4 M and \alpha = ?

K_a=5.8\times 10^{-10}

Putting in the values we get:

5.8\times 10^{-10}=\frac{(4.4\times \alpha)^2}{(4.4-4.4\times \alpha)}

(\alpha)=0.000011

[H^+]=c\times \alpha

[H^+]=4.4\times 0.000011=4.8\times 10^{-5}M

Also pH=-log[H^+]

pH=-log[4.8\times 10^{-5}]=4.3

Thus pH of a 4.4 M H_3BO_3 solution is 4.3

3 0
3 years ago
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