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Alex777 [14]
3 years ago
6

Select the reagent that will separate the pairs of ions listed.

Chemistry
1 answer:
daser333 [38]3 years ago
8 0

1 - a, 2-b, 3-d, 4 - e

<u>Explanation:</u>

<u></u>A g^{+} \text {and } A l^{3+} - Separated by 6M HCl

Silver and aluminum ions can be separated using the acid, 6M HCl.

C u^{2+} \text { and } C a^{2+}- Separated by H₂S

Copper and Calcium ions can be separated by the reagent hydrogen sulfide, H₂S.

M g^{2+} \text { and } K^{+} - Separated by (NH₄)₂HPO₄ in NH₃

Magnesium and Potassium ions can be separated by the reagent Ammonium hydrogen phosphate in ammonia.

N i^{2+} \text { and } C o^{2+}- cannot be separated

Nickel and Cobalt ion cannot be separated by the reagents given at all.

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Decomposition:chemical change whereby a molecule breaks down into simpler moles clues or elements
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A 50.6 grams sample of magnesium hydroxide (Mg(OH)2) is reacted with 45.0 grams of hydrochloric acid (HCl). What is the theoreti
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grams HCl required = (50.6 grams Mg(OH)2) * (1 mol Mg(OH)2 / 58.3197 grams Mg(OH)2) * (2 mol HCl / 1 mol Mg(OH)2) * (36.453 grams HCl / 1 mol HCl) = 63.26 grams HCl required

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theoretical yield MgCl2 = (45.0 grams HCl) * (1 mol HCl / 36.453 grams HCl) * (1 mol MgCl2 / 2 mol HCl) * (95.211 grams MgCl2 / 1 mol MgCl2) = 58.6 grams MgCl2 </span>
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Be sure to answer all parts. Styrene is produced by catalytic dehydrogenation of ethylbenzene at high temperature in the presenc
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Answer:

a) ΔHºrxn = 116.3 kJ, ΔGºrxn = 82.8 kJ,  ΔSºrxn =  0.113 kJ/K

b) At 753.55 ºC or higher

c )ΔG =  1.8 x 10⁴ J

    K = 8.2 x 10⁻²

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a)                                 C6H5−CH2CH3  ⇒  C6H5−CH=CH2  + H₂

ΔHf kJ/mol                    -12.5                           103.8                      0

ΔGºf kJ/K                        119.7                         202.5                      0

Sº J/mol                          255                          238                      130.6*

Note: This value was not given in our question, but is necessary and can be found in standard handbooks.

Using Hess law to calculate  ΔHºrxn we have

ΔHºrxn  = ΔHfº C6H5−CH=CH2 +  ΔHfº H₂ - ΔHºfC6H5−CH2CH3

ΔHºrxn =     103.8 kJ + 0 kJ  - (-12.5 kJ)

ΔHºrxn = 116.3 kJ

Similarly,

ΔGrxn = ΔGºf C6H5−CH=CH2 +  ΔGºfH₂ - ΔGºfC6H5CH2CH3

ΔGºrxn=   202.5 kJ + 0 kJ - 119.7 kJ  = 82.8 kJ

ΔSºrxn = 238 J/mol + 130.6 J/mol -255 J/K = 113.6 J/K = 0.113 kJ/K

b) The temperature at which the reaction is spontaneous or feasible occurs when ΔG becomes negative and using

ΔGrxn =  ΔHrxn -TΔS

we see that will happen when the term  TΔS  becomes greater than ΔHrxn since ΔS  is positive  , and so to sollve for T we will make ΔGrxn equal to zero and solve for T. Notice here we will make the assumption that  ΔºHrxn and ΔSºrxn remain constant at the higher temperature  and will equal the values previously calculated for them. Although this assumption is not entirely correct, it can be used.

0 = 116 kJ -T (0.113 kJ/K)

T = 1026.5 K  =  (1026.55 - 273 ) ºC = 753.55 ºC

c) Again we will use

                       ΔGrxn =  ΔHrxn -TΔS

to calculate ΔGrxn   with the assumption that ΔHº and ΔSºremain constant.

ΔG =  116.3 kJ - (600+273 K) x 0.113 kJ/K =  116.3 kJ - 873 K x 0.113 kJ/K

ΔG =  116.3 kJ - 98.6 kJ =  17.65 kJ = 1.8 x 10⁴ J ( Note the kJ are converted to J to necessary for the next part of the problem )

Now for solving for K, the equation to use is

ΔG = -RTlnK and solve for K

- ΔG / RT = lnK  ∴ K = exp (- ΔG / RT)

K = exp ( - 1.8 x 10⁴ J /( 8.314 J/K  x 873 K)) = 8.2 x 10⁻²

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