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KiRa [710]
3 years ago
13

What is the study of the cryosphere called?

Chemistry
1 answer:
SVEN [57.7K]3 years ago
6 0
The study of the Cryosphere is called:

Cryology 

I hope this helps! :)

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Study the following reaction carefully. What classification should this reaction have? 4Al + 3O 2 2Al 2 O 3
nikitadnepr [17]

The reaction is a synthesis…

4Al + 3O_2 \longrightarrow 2Al_2O_3\\

… because <em>two substances are combining to make one other substance</em>.

4\stackrel{\hbox{0}}{\hbox{Al}} + 3\stackrel{\hbox{0}}{\hbox{O}}_2 \longrightarrow 2\stackrel{\hbox{+3}}{\hbox{Al}}_2\stackrel{\hbox{-2}}{\hbox{O}}_3\\

It is also a <em>reduction-oxidation reaction</em> because the <em>oxidation number of Al increases</em> from 0 to +3 and the <em>oxidation number of O</em> decreases from 0 to -2.

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Kc for the reaction of hydrogen and iodine to produce hydrogen iodide, H2(g) I2(g) ⇌ 2HI(g) is 54.3 at 430°C. Determine the init
Jlenok [28]

Answer : The initial concentration of HI and concentration of HI at equilibrium is, 0.27 M and 0.386 M  respectively.

Solution :  Given,

Initial concentration of H_2 and I_2 = 0.11 M

Concentration of H_2 and I_2 at equilibrium = 0.052 M

Let the initial concentration of HI be, C

The given equilibrium reaction is,

                          H_2(g)+I_2(g)\rightleftharpoons 2HI(g)

Initially               0.11   0.11            C

At equilibrium  (0.11-x) (0.11-x)   (C+2x)

As we are given that:

Concentration of H_2 and I_2 at equilibrium = 0.052 M  = (0.11-x)

0.11 - x = 0.052

x = 0.11 - 0.052

x = 0.058 M

The expression of K_c will be,

K_c=\frac{[HI]^2}{[H_2][I_2]}

54.3=\frac{(C+2(0.058))^2}{(0.052)\times (0.052)}

By solving the terms, we get:

C = 0.27 M

Thus, initial concentration of HI = C = 0.27 M

Thus, the concentration of HI at equilibrium = (C+2x) = 0.27 + 2(0.058) = 0.386 M

3 0
3 years ago
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