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Darya [45]
3 years ago
11

Which of the following processes have a ΔS < 0? Which of the following processes have a ΔS < 0? carbon dioxide(g) → carbon

dioxide(s) water freezes propanol (g, at 555 K) → propanol (g, at 400 K) methyl alcohol condenses All of the above processes have a ΔS < 0.
Chemistry
1 answer:
Umnica [9.8K]3 years ago
6 0

Answer:

All of the above processes have a ΔS < 0.

Explanation:

ΔS represents change in entropy of a system. Entropy refers to the degree of disorderliness of a system.

The question requests us to identify the process that has a negative change of entropy.

carbon dioxide(g) → carbon dioxide(s)

There is  a change in state from gas to solid. Solid particles are more ordered than gas particles so this is a negative change in entropy.

water freezes

There is  a change in state from liquid to solid. Solid particles are more ordered than liquid particles so this is a negative change in entropy.

propanol (g, at 555 K) → propanol (g, at 400 K)

Temperature is directly proportional to entropy, this means higher temperature leads t higher entropy.

This reaction highlights a drop in temperature which means a negative change in entropy.

methyl alcohol condenses

Condensation is the change in state from gas to liquid. Liquid particles are more ordered than gas particles so this is a negative change in entropy.

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The change in the velocity = 4 m/s

Acceleration = 4 m/s²

<h3>Further explanation</h3>

Given

vo = initial velocity = 4 m/s

vf = final velocity = 8 m/s

t = 1 s

Required

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Acceleration

Solution

the change in velocity =

\tt vf-vo=8-4=4~m/s

Acceleration = ratio of a change in velocity and the time

\tt a=\dfrac{\Delta v}{t}= \dfrac{vf-vo}{t}

Input the value :

\tt a=\dfrac{4~m/s}{1~s}=4~m/s^2

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A volume of 105 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If
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Answer:

<h3>25.0 grams is the mass of the steel bar.</h3>

Explanation:

Heat gained by steel bar will be equal to heat lost by the water

Q_1=-Q_2

Mass of steel= m_1

Specific heat capacity of steel = c_1=0.452 J/g^oC

Initial temperature of the steel = T_1=2.00^oC

Final temperature of the steel = T_2=T=21.50^oC

Q_1=m_1c_1\times (T-T_1)

Mass of water= m_2= 105 g

Specific heat capacity of water=c_2=4.18 J/g^oC

Initial temperature of the water = T_3=22.00^oC

Final temperature of water = T_2=T=21.50^oC

Q_2=m_2c_2\times (T-T_3)-Q_1=Q_2(m_1c_1\times (T-T_1))=-(m_2c_2\times (T-T_3))

On substituting all values:

(m_1\times 0.452 J/g^oC\times (21.50^o-2.00^oC))=-(105 g\times 4.18 J/g^oC\times (21.50^o-22.00^o))\\\\m_1*8.7914=241.395\\\\m_1=\frac{219.45}{8.7914} \\\\m_1=24.9\\\\ \approx25 \texttt {grams}

<h3>25.0 grams is the mass of the steel bar.</h3>
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Answer:

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