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Alenkasestr [34]
3 years ago
10

Calculate the heat energy released when 13.3 g of liquid mercury at 25.00 C is converted to solid mercury at its melting point.C

onstants for mercury at 1 atmheat capacity of Hg(l) 28.0 J/(mol K)melting point 234.32 Kenthalphy of fusion 2.29 KJ/mol
Physics
1 answer:
dmitriy555 [2]3 years ago
4 0

Answer:

-270.321012\ J

Explanation:

C_v = Heat capacity of Hg = 28 J/mol

\Delta T = Change in temperature = (234.32-(273.15+25))

\Delta H_{f} = Enthalpy of fusion = 2.29 kJ/mol

The number of moles is given by

n=13.3\times \dfrac{1}{200.59}\\\Rightarrow n=0.0663\ molHg

Heat is given by

Q_1=nC_v\Delta T\\\Rightarrow Q_1=0.0663\times 28\times (234.32-(273.15+25))\\\Rightarrow Q_1=-118.494012\ J

Heat released is given by

Q_2=-n\Delta H_{f}\\\Rightarrow Q_2=-0.0663\times 2.29\times 10^3\\\Rightarrow Q_2=-151.827\ J

Total heat is given by

Q=Q_1+Q_2\\\Rightarrow Q=-118.494012+(-151.827)\\\Rightarrow Q=-270.321012\ J

The total heat released is -270.321012\ J

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<h3>Motion Under Gravity</h3>

The motion of an object under gravity is the vertical motion of the object under the influence of acceleration due to gravity.

Given that a ball is thrown horizontally from the roof of a building 75 m tall with a speed of 4.6 m/s.

a. how much later does the ball hit the ground?

The time can be calculated by considering the vertical component of the motion with the use of formula below.

h = ut + 1/2gt²

Where

  • Height h = 75 m
  • Initial velocity u = 0 ( vertical velocity )
  • Acceleration due to gravity g = 9.8 m/s²
  • Time t = ?

Substitute all the parameters into the formula

75 = 0 + 1/2 × 9.8 × t²

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t = 3.9 s

b. how far from the building will it land?

The range can be found by using the formula

R = ut

Where u = 4.6 m/s ( horizontal velocity )

R = 4.6 × 3.9

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c. what is the velocity of the ball just before it hits the ground?

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v = u + gt

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Therefore, the answers are 3.9 s, 18 m and 42.82 m/s

Learn more about Vertical motion here: brainly.com/question/24230984

#SPJ1

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