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3 years ago

Calculate the amount of heat liberated (in kj) from 411 g of mercury when it cools from 88.0°c to 12.0°c.

1 answer:

5 0

You should note that the melting point of mercury is -38.83°C, while the boiling point is at 356.7°C. Then, that means that there is no latent heat involved here. We only compute for the sensible heat.

ΔH = mCpΔT
The Cp of mercury is 0.14 J/g·°C
Thus,
ΔH = (411 g)(0.14 J/g·°C)(88 - 12°C)
<em>ΔH = 4,373.04 J</em>

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Acid rain is formed when chemicals in the air get into rain and up the acidity levels.

tester [92]

False! Because the acid rain it becomes a chemicals it does not air get into the rain and up the acidity levels.
Hope it helped you.

-Charlie

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3 years ago

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Relating to Newton’s Third Law, action and reaction, what is the reaction when a rocket expels gas, smoke and flames from the no

Cloud [144]

Answer:

Propels in the opposite direction

Explanation:

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2 years ago

A hopper jumps straight up to a height of 1.3 m. With what velocity did he leave the floor

xxMikexx [17]

The velocity with which the jumper leaves the floor is 5.1 m/s.

<h3>What is the initial velocity of the jumper?</h3>

The initial velocity of the jumper or the velocity with which the jumper leaves the floor is calculated by applying the principle of conservation of energy as shown below.

Kinetic energy of the jumper at the floor = Potential energy of the jumper at the maximum height

¹/₂mv² = mgh

v² = 2gh

v = √2gh

where;

v is the initial velocity of the jumper on the floor
h is the maximum height reached by the jumper
g is acceleration due to gravity

v = √(2 x 9.8 x 1.3)

v = 5.1 m/s

Learn more about initial velocity here: brainly.com/question/19365526
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8 months ago

The nuclei of large atoms, such as uranium, with 9292 protons, can be modeled as spherically symmetric spheres of charge. The ra

Scrat [10]

Answer:

Part 1 $E = 2.42 * 10^{21} N/C$

Part 2 $E = 1.3 * 10^{13} N/C$

Part 3 $E = 0$

Explanation:

Given

Number of protons $= 92$

Radius of nucleus $r_n = 7.4 * 10^{-15} m$

Distance of the electrons $r_1 = 1.0 * 10^ {-10} m$

Part 1

Electric field produced by just outside its surface

$E = \frac{q}{4\pi*E_0* r_n^2 } \\E = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(7.4* 10^{-15})^2} \\E = 2.42 * 10^{21} N/C$

Part 2

Electric field produced by just outside its surface

$E = \frac{q}{4\pi*E_0* r_n^2 } \\E = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(1* 10^{-15})^2} \\E = 1.3 * 10^{13} N/C$

Part 3

The net electric field inside a uniform shell of negative charge is zero because the electric flux lines cancel out each other

hence, the solution is

Part 1 $E = 2.42 * 10^{21} N/C$

Part 2 $E = 1.3 * 10^{13} N/C$

Part 3 $E = 0$

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2 years ago

a large heavy truck and a baby carriage roll down a hill. Neglecting friction, at the bottom of the hill, the baby carriage will

Lisa [10]

Answer:

please give me brainlist and follow

Explanation:

At the bottom of the hill, the baby carriage will likely have less momentum Therefore, option D is correct. Solution: ... Therefore, at the bottom of the hill, the heavy truck will have more momentum and baby carriage will have less momentum.

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3 years ago