HURRY ILL MARK BRAINLIEST

An object at 20∘C absorbs 25.0 J of heat. What is the change in entropy ΔS of the object? Express your answer numerically in jou

nevsk [136]

Question:

Part D) An object at 20∘C absorbs 25.0 J of heat. What is the change in entropy ΔS of the object? Express your answer numerically in joules per kelvin.

Part E) An object at 500 K dissipates 25.0 kJ of heat into the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.

Part F) An object at 400 K absorbs 25.0 kJ of heat from the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.

Part G) Two objects form a closed system. One object, which is at 400 K, absorbs 25.0 kJ of heat from the other object,which is at 500 K. What is the net change in entropy ΔSsys of the system? Assume that the temperatures of the objects do not change appreciably in the process. Express your answer numerically in joules per kelvin.

Answer:

D) 85 J/K

E) - 50 J/K

F) 62.5 J/K

G) 12.5 J/K

Explanation:

Let's make use of the entropy equation: ΔS = $\frac{Q}{T}$

Part D)

Given:

T = 20°C = 20 +273 = 293K

Q = 25.0 kJ

Entropy change will be:

ΔS = $\frac{25*1000}{293}$

= 85 J/K

Part E)

Given:

T = 500K

Q = -25.0 kJ

Entropy change will be:

ΔS = $\frac{-25*1000}{500}$

= - 50 J/K

Part F)

Given:

T = 400K

Q = 25.0 kJ

Entropy change will be:

ΔS = $\frac{25*1000}{400}$

= 62.5 J/K

Part G:

Given:

T1 = 400K

T2 = 500K

Q = 25.0 kJ

The net entropy change will be:

ΔS = $(\frac{25*1000}{400}) + (\frac{-25*1000}{500}$

= 12.5 J/K

7 0

3 years ago

The half-life of gold-198 is 2.7 days. After

Pie

Answer: 8.1 days

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant = x

a - x = amount left after decay process= $\frac{x}{4}$

a) to find rate constant

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

$t_{\frac{1}{2}}=\frac{0.693}{k}$

$k=\frac{0.693}{2.7days}=0.257days^{-1}$

b) for completion of one fourth of reaction

$t=\frac{2.303}{k}\log\frac{x}{\frac{x}{4}}$

$t=\frac{2.303}{0.257}\log{4}$

$t=8.1days$

Thus after 8.1 days , one fourth of original amount will remain.

8 0

4 years ago

What is the mass of 0.2 moles of ethanol

Verdich [7]

Answer:

Answer is 9.2 grams

Explanation:

Molar mass of ethanol

${ \bf{CH _{3}CH_{2}OH}} = (12 \times 2) + (6 \times 1) + (16) \\ = 46 \: g$

for moles:

${ \sf{1 \: mole \: of \: ethanol \: weighs \: 46 \: g}} \\ { \sf{0.2 \: moles \: weigh \: (0.2 \times 46) \: g}} \\ = 9.2 \: g$

4 0

3 years ago

List the steps of the scientific method

NNADVOKAT [17]

1. Make an Observation. Scientists are naturally curious about the world. ...
2. Form a Question. After making an interesting observation, a scientific mind itches to find out more about it. ...
3. Form a Hypothesis. ...
4. Conduct an Experiment. ...
5. Analyse the Data and Draw a Conclusion.

5 0

3 years ago

Which subatomic particle can vary between NEUTRAL atoms of the SAME element

tekilochka [14]

Answer:

answer is-neutrons

Explanation:

fj’yf’jg’kh’jy’tfthfy,gtj ftjfrvf

7 0

3 years ago