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pashok25 [27]
3 years ago
7

A sample of gas in which [h2s] = 5.25 m is heated to 1400 k in a sealed vessel. after chemical equilibrium has been achieved, wh

at is the value of [h2s]? assume no h2 or s2 was present in the original sample.
Chemistry
1 answer:
raketka [301]3 years ago
4 0
When the reaction equation is:

by using the ICE equation:

                   2H2S(g)  ↔ 2 H2(g) +  S2(s)

initial            5.25                 0            0

change         -2X                 +2X         +X

Equ           (5.25-2X)              2X           X

and when Kc = 2.2 x 10^-4 

Kc = [H2]^2/ [H2S]^2

    = (2x)^2 / (5.25-2X)^2 by solving for X 

∴X = 0.038 M

∴[H2S] = 5.25 - 2X 

             = 5.25 - (2*0.038)

            = 5.174 M

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Answer:

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Explanation:

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In the open chain form, monosaaccharides have a carbonuyl group in one of their chains. If the carbonyl group is in the form of an aldehyde group, the monosaccharide is an aldose; if the carbonyl group is in the form of a ketone group, the monosaccharide is known as a ketose. glucose is an aldose while fructose is a ketose.

In D-glucose, there is an aldehyde functional group, and the carbonyl group is at carbon-1 when looking at the Fischer projection.

In D-fructose, there is a ketone functional group, and the carbonyl group is at carbon-2 when looking at the Fischer projection.

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3 years ago
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Explanation:

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In a titration, for an acid to neutralize a base, at the equivalence point, there should be an equal number of moles of H+ and OH-.

Moles of OH- can be found by multiplying the concentration of the base by the volume. (You will need to keep in mind the stoichimetric coefficients if the strong base is Ca(OH)₂, Ba(OH)₂, or Sr(OH)₂.

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