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sleet_krkn [62]
2 years ago
6

A 0.200-g sample of cobalt metal reacted with hydrochloric acid according to the following balanced chemical equation: Co (s) +

2 HCl (aq) ➝ CoCl2 (aq) + H2 (g) The volume of hydrogen gas collected over water was 87.5 mL at 20 °C and a barometer reading of 763 mm Hg. Referring to Example Exercise 1, the adjusted volume of the H2 gas is
Chemistry
1 answer:
eimsori [14]2 years ago
3 0

Answer:

V H2(g) = 83.13 mL

Explanation:

Dalton's law:

  • Pgas = Pbar - P*H2O

∴ vapor pressure (P*)

⇒ P*H2O(20°C) = 17.5 mmHg

∴ Pbar = 763 mmHg

⇒ PH2 = 763 mmHg - 17.5 mmHg = 745.5 mmHg = 0.9809 atm

  • Co(s) + 2 HCl(aq) → CoCl2(aq) + H2(g)

∴ Mw Co(s) = 58.933 g/mol

⇒ mol Co(s) = ( 0.200 g Co )×( mol/58.933 g ) = 3.394 E-3 mol Co

⇒ n H2(g) = ( 3.394 E-3 mol Co )( mol H2 / mol Co ) = 3.394 E-3 mol H2(g)

ideal gas:

  • PV = RTn

∴ R = 0.082 atm.L/K.mol

∴ T = 20°C ≅ 293 K

the adjusted volume H2(g):

⇒ V H2(g) = ((0.082 atm.L/K.mol)(293 K)(3.394 E-3 mol)) / (0.9809 atm)

⇒ V H2(g) = 0.08313 L

⇒ V H2(g) = 83.13 mL

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<span>Answer: A 1.00 L solution containing 3.00x10^-4 M Cu(NO3)2 and 2.40x10^-3 M ethylenediamine (en). contains 0.000300 moles of Cu(NO3)2 and 0.00240 moles of ethylenediamine by the formula Cu(en)2^2+ 0.000300 moles of Cu(NO3)2 reacts with twice as many moles of en = 0.000600 mol of en so, 0.00240 moles of ethylenediamine - 0.000600 mol of en reacted = 0.00180 mol en remains by the formula Cu(en)2^2+ 0.000300 moles of Cu(NO3)2 reacts to form an equal 0.000300 moles of Cu(en)2^2+ Kf for Cu(en)2^2+ is 1x10^20. so 1 Cu+2 & 2 en --> Cu(en)2^2+ Kf = [Cu(en)2^2+] / [Cu+2] [en]^2 1x10^20. = [0.000300] / [Cu+2] [0.00180 ]^2 [Cu+2] = [0.000300] / (1x10^20) (3.24 e-6) Cu+2 = 9.26 e-19 Molar since your Kf has only 1 sig fig, you might be expected to round that off to 9 X 10^-19 Molar Cu+2</span>
4 0
2 years ago
Gamma rays are often used to kill microorganisms in food, in an attempt to make the food safer. Some people contend that this ir
nikdorinn [45]

Answer:

b . Irradiated food is shown to not be radioactive.

Explanation:

If it can be proven that irradiated food is not radioactive, then it will effective dispute the idea that irradiated food are less safe to eat.

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  • Thus, bacteria and other food spoilers can be exterminated from the food.
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If it can be proven, that this is true, then it will challenge the idea that irradiated foods are not safe.

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1 year ago
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2 years ago
Be sure to answer all parts. For each reaction, find the value of ΔSo. Report the value with the appropriate sign. (a) 3 NO2(g)
Liono4ka [1.6K]

Answer:

a. -268.13 J/K

b. -279.95 J/K

c. + 972.59 J/K

Explanation:

The value of the change in entropy (ΔS°) can be calculated by:

ΔS° = ∑n*S° products - ∑n*S° reactants, where n is the stoichiometric number of moles.

The values of S° for each substance can be found on a thermodynamic table.

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S°, NO2(g) = 240.06 J/mol.K

S°, H2O(l) = 69.91 J/mol.K

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S°, NO(g) = 210.76 J/mol.K

ΔS° = (210.76 + 2*155.60) - (3*240.06 + 69.91)

ΔS° = -268.13 J/K

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Since the cost of materials cannot be affected and have to be bought under similar rate of cost of working or process can be decreased. By using  machines calculation and mechanical devices this come would be accurate and the laws would be minimised. Hence the companies outcome would increase.

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2 years ago
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