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Gelneren [198K]
2 years ago
9

A chunk of ice breaks off a glacier and falls 30.0 m before it hits the water. Assuming it falls freely (there is no air resista

nce), how long does it take to hit the water? Which quantity increases faster, the speed of the ice chunk or its distance traveled?
Chemistry
1 answer:
lina2011 [118]2 years ago
7 0

Answer:

The ice chunk will take 2.47 seconds to hit the water.

Explanation:

Height from which ice chunk fell = h = 30.0 m

Initial velocity of the ice chunk = u = 0 m/s

Time taken by ice chunk to cover 30.0 m of height = t

Acceleration due to gravity = g 9.8m/s^2

Using second equation of motion ;

h=ut+\frac{1}{2}gt^2

30.0 m=0 m/s\times t+\frac{1}{2}\times 9.8 m/s^2\times t^2

t=\sqrt{\frac{30.0m/s\times 2}{9.8 m/s^2}}2.47 s

The ice chunk will take 2.47 seconds to hit the water.

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Explanation:

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- 10.555 kJ/mol.

Explanation:

∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.

Where, ∆G°rxn is the standard free energy change of the reaction (J/mol).

∆H°rxn is the standard enthalpy change of the reaction (J/mol).

T is the temperature of the reaction (K).

∆S°rxn is the standard entorpy change of the reaction (J/mol.K).

  • Calculating ∆H°rxn:

∵ ∆H°rxn = ∑∆H°products - ∑∆H°reactants

<em>∴ ∆H°rxn = (2 x ∆H°f NOCl) - (1 x ∆H°f Cl₂) - (2 x ∆H°f NO) </em>= (2 x 51.71 kJ/mol) - (1 x 0) - (2 x 90.29 kJ/mol) = - 77.16 kJ/mol.

  • Calculating ∆S°rxn:

∵  ∆S°rxn = ∑∆S°products - ∑∆S°reactants

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<em></em>

  • Calculating ∆G°rxn:

∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.

<em>∴ ∆G°rxn = ∆H°rxn - T∆S°rxn </em>= (- 77.16 kJ/mol) - (550 K)(- 0.1211 kJ/mol.K) = <em>- 10.555 kJ/mol.</em>

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3 years ago
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