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photoshop1234 [79]
3 years ago
6

What is the value of the equilibrium constant for the reaction between Cr(s) and Cu^2+ (aq) at 25 degree C? Cr(s) + Cu^2+ (q)rig

htarrow Cr^3+ (aq) + Cu(s) If your calculator has overflow issues with this problem, you can use this identity to solve for K: if log(K) = x + y then K = 10^x times 10^y 3 times 10^109 -6 times 10^5 2 times 10^18 1 times 10^2 3 times 10^40
Chemistry
1 answer:
Ludmilka [50]3 years ago
3 0

Answer:

3 × 10¹⁰⁹

Explanation:

2Cr (s) → 2Cr³⁺ + 6e⁻        

and,

Ec° for the above reaction = 0.74 V

also,

3Cu²⁺ (aq) + 6e⁻ →   3Cu(s)

and,

Ea° for the above reaction = 0.34 V

Now,

The cell potential, E° = 0.34 + 0.74 = 1.08

E° = \frac{\textup{0.059}}{\textup{n}}\log k

thus,

0.34 = \frac{\textup{0.059}}{\textup{n}}\log k

here, n is the number of electrons exchanged

n = 6

1.08 = \frac{\textup{0.05921}}{\textup{6}}\log k

or

㏒ k = 109.44

taking anti-log both sides

we get

k =  3 × 10¹⁰⁹

Hence, the answer is option (A) 3 × 10¹⁰⁹

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How many atoms are there in 1 g of argon?
amm1812

Answer:

1.5057×10^22 atom

Explanation:

As we

1 mole of argon = 40 g of argon

i.e 40 g of argon = 1 mole of argon

1 g of argon = 1/40 mole of argon

1 mole of argon = 6.023×10^23 atom of argon

1/40 mole if argon = 1/40 ×6.023×10^23

= 1.5057×10^22

7 0
3 years ago
Why is the energy supplied by the cooker greater than that calculated ?
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Answer:

Explanation:

Q1.

(a) 46 200

accept 46 000

allow 1 mark for correct substitution

ie 0.5 × 4200 × 22 provided no subsequent step

2

(b) Energy is used to heat the kettle.

[3]

Q2.

(a) (approximate same size particles as each other and as liquid and gas) touching

do not accept particles that overlap

regular arrangement (filling the square)

(b) condensing

(c) solid

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(f) mass of the liquid

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672 000 (J)

an answer of 672 000 (J) scores 2 marks

[11]

Q3.

(a) x-axis labelled and suitable scale

Page 12 of 13

points plotted correctly

allow 5 correctly plotted for 2 marks, 3−4

correctly plotted for 1 mark

allow ± ½ square

2

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(b)

allow ecf from line of best fit in part (a)

0.075 (°C/s)

an answer of 0.075 (°C/s) scores 2 marks

(c) Δθ = 11.5 (°C)

a calculation using an incorrect temperature

scores max 3 marks

ΔE = 1.50 × 900 × 11.5

ΔE = 15 525 (J)

ΔE = 15.525 (kJ)

an answer of 15.525 (kJ) or 15.53 (kJ) or 15.5

(kJ) scores 4 marks

an answer of 15 525 (kJ) scores 3 marks

[10]

Q4.

(a) 80 °C

ΔE = 0.5 × 3400 × 80

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an answer of 136 000 (J) scores 3 marks

(b) energy is dissipated into the surroundings

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(c) put a lid on the pan

allow any sensible practical suggestion

eg add salt to the water

Page 13 of 13

(d) efficiency = 300/500

efficiency = 0.6

an answer of 0.6 or 60% scores 2 marks

allow efficiency = 60%

an answer of 0.6 with a unit scores 1 mark

an answer of 60 without a unit scores 1 mark

7 0
2 years ago
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A) Head to tail joining of monomers. :) (confirmed correct answer, I took the test)

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