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Readme [11.4K]
3 years ago
8

An electroplating solution is made up of nickel(II) sulfate. How much time would it take to deposit 0.500 g of metallic nickel o

n a custom car part using a current of 3.00 A
Physics
1 answer:
ELEN [110]3 years ago
6 0

Answer:

1.52 hour

Explanation:

M = 0.5 g, I = 3 A

Electrochemical equivalent of nickel

Z = 3.04 × 10^(-4) g/C

By use of Faraday's laws of electrolysis

M = Z I t

t = M / Z I

t = 0.5 / (3.04 × 10^-4 × 3)

t = 5482.45 second = 1.52 hour

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What change in entropy occurs when a 0.15 kg ice cube at -18 °C is transformed into steam at 120 °c 4.
Studentka2010 [4]

<u>Answer:</u> The change in entropy of the given process is 1324.8 J/K

<u>Explanation:</u>

The processes involved in the given problem are:

1.)H_2O(s)(-18^oC,255K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\3.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\4.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\5.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(120^oC,393K)

Pressure is taken as constant.

To calculate the entropy change for same phase at different temperature, we use the equation:

\Delta S=m\times C_{p,m}\times \ln (\frac{T_2}{T_1})      .......(1)

where,

\Delta S = Entropy change

C_{p,m} = specific heat capacity of medium

m = mass of ice = 0.15 kg = 150 g    (Conversion factor: 1 kg = 1000 g)

T_2 = final temperature

T_1 = initial temperature

To calculate the entropy change for different phase at same temperature, we use the equation:

\Delta S=m\times \frac{\Delta H_{f,v}}{T}      .......(2)

where,

\Delta S = Entropy change

m = mass of ice

\Delta H_{f,v} = enthalpy of fusion of vaporization

T = temperature of the system

Calculating the entropy change for each process:

  • <u>For process 1:</u>

We are given:

m=150g\\C_{p,s}=2.06J/gK\\T_1=255K\\T_2=273K

Putting values in equation 1, we get:

\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K

  • <u>For process 2:</u>

We are given:

m=150g\\\Delta H_{fusion}=334.16J/g\\T=273K

Putting values in equation 2, we get:

\Delta S_2=\frac{150g\times 334.16J/g}{273K}\\\\\Delta S_2=183.6J/K

  • <u>For process 3:</u>

We are given:

m=150g\\C_{p,l}=4.184J/gK\\T_1=273K\\T_2=373K

Putting values in equation 1, we get:

\Delta S_3=150g\times 4.184J/g.K\times \ln(\frac{373K}{273K})\\\\\Delta S_3=195.9J/K

  • <u>For process 4:</u>

We are given:

m=150g\\\Delta H_{vaporization}=2259J/g\\T=373K

Putting values in equation 2, we get:

\Delta S_2=\frac{150g\times 2259J/g}{373K}\\\\\Delta S_2=908.4J/K

  • <u>For process 5:</u>

We are given:

m=150g\\C_{p,g}=2.02J/gK\\T_1=373K\\T_2=393K

Putting values in equation 1, we get:

\Delta S_5=150g\times 2.02J/g.K\times \ln(\frac{393K}{373K})\\\\\Delta S_5=15.8J/K

Total entropy change for the process = \Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4+\Delta S_5

Total entropy change for the process = [21.1+183.6+195.9+908.4+15.8]J/K=1324.8J/K

Hence, the change in entropy of the given process is 1324.8 J/K

4 0
3 years ago
engineers are using computer models to study train collisions design safer train cars. They start by modeling an elastic collisi
Archy [21]

Answer:

The final velocity of the second car is 57 m/s south.

Explanation:

This is an elastic collision between two train cars. In this case, the total kinetic energy between the two bodies will remain the same.

The formula to apply is :

m_1v_1i +m_2v_2i=m_1v_1f+m_2v_2f

where ;

m_1=mass of object 1\\v_1i=initial  velocity object1\\m_2=mass of object2\\v_2i=initial velocity object 2\\v_1f=final velocity object 1\\v_2f=final velocity object 2

Given in the question that;

m_1=14650kg\\v_1i=18m/s\\m_2=3825kg\\v_2i=11m/s\\v_1f=6m/s\\v_2f=?

Apply the formula as;

m_1v_1i +m_2v_2i=m_1v_1f+m_2v_2f

{14650*18}+{3825*11} = {14650 *6} + {3825 * v₂f}

263700+42075=87900 + 3825v₂f

305775 =87900 + 3825v₂f

305775-87900 = 3825v₂f

217875=3825v₂f

217875/3825 =v₂f

56.96 = v₂f

<u>57 m/s = v₂f { nearest whole number}</u>

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2 years ago
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The unconscious mind plays a large role in Sigmund Freud's psychoanalysis<span>. He discussed the importance of the unconscious mind in understanding conscious thought and behaviour.   </span><span>Sigmund Freud

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How long do molecules of groundwater stay in the ground?
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CORRECT ANSWER:

d. Anywhere from days to thousands of years.

STEP-BY-STEP EXPLANATION:

The whole question from book is

How long do molecules of groundwater stay in the

ground?

a. Days

b. Weeks

c. Months

d. Anywhere from days to thousands of years

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