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18 Oct 2016, 00:55
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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79% (01:05) correct
21%
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based on 149
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It is well known that the ratio of hydrogen’s mass to oxygen’s mass is 1 to 16. If the moving speed rates of hydrogen atoms and oxygen atoms are inversely proportional to square root of its mass, what is the ratio of the moving speed rate of hydrogen to that of oxygen?
A. 1:4
B. 4:1
C. 1:16
D. 16:1
E. 1:2
* A solution will be posted in 2 days.
A. 1:4
B. 4:1
C. 1:16
D. 16:1
E. 1:2
* A solution will be posted in 2 days.
18 Oct 2016, 03:09
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It is well known that the ratio of hydrogen’s mass to oxygen’s mass is 1 to 16. If the moving speed rates of hydrogen atoms and oxygen atoms are inversely proportional to square root of its mass, what is the ratio of the moving speed rate of hydrogen to that of oxygen?
A. 1:4
B. 4:1
C. 1:16
D. 16:1
E. 1:2
Given ratio of M(H) / M(O) = 1/16 where...M(H) - hydrogen mass and M(O) - Oxygen mass.
Given S(H) = K/√M(H) and S(O) = K/√M(O) - Where K is constant and since they are inversely proportional we can write in above form.
S(H) = K/√M(H) => S(H) = K/√1=> S(H) = K/1
S(O) = K/√M(O) => S(O) = K/√16 => S(O) = K/4
Now We need to find S(H)/S(O) => K/1 / K / 4 => 4:1.
Hence option B.
A. 1:4
B. 4:1
C. 1:16
D. 16:1
E. 1:2
Given ratio of M(H) / M(O) = 1/16 where...M(H) - hydrogen mass and M(O) - Oxygen mass.
Given S(H) = K/√M(H) and S(O) = K/√M(O) - Where K is constant and since they are inversely proportional we can write in above form.
S(H) = K/√M(H) => S(H) = K/√1=> S(H) = K/1
S(O) = K/√M(O) => S(O) = K/√16 => S(O) = K/4
Now We need to find S(H)/S(O) => K/1 / K / 4 => 4:1.
Hence option B.
18 Oct 2016, 09:17
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MathRevolution
By Mass : \(\frac{Hydrogen}{Oxygen}\) = \(\frac{1}{16}\)
By moving speed rate : \(\frac{Hydrogen}{Oxygen}\) = \(\sqrt{\frac{1}{16}}\) = \(\frac{1}{4}\)
Quote:
\(\frac{Moving \ speed \ rate}{Mass \ rate}\) = \(4:1\)
Hence answer will be (B) \(4:1\)
