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3 years ago

Predict which molecule will show a similar relationship between its heat of fusion and its heat vaporization that water does

Physics

2 answers:

BaLLatris [955]3 years ago

8 0

Question:

A) C6H6

B) CH3CH2CH2CH2CH2COH6

C) NaCl

D) NH3

Answer:

The correct option is;

A) C₆H₆

Explanation:

Heat of fusion = 6.02 kJ/mol

Heat of vaporization =40.8 kJ/mol

Here, we analyze each of the options as follows

A) C₆H₆

Benzene has a melting point of 5.5° C and a boiling point of

80.1 ° C similar to water

Heat of fusion = 9.92 kJ/mol

Heat of vaporization =30.8kJ/mol

B) CH₃CH₂CH₂CH₂CH₂COH₆

The above compound is more likely solid

C) NaCl solid

D) NH₃ melting point = -77.73 °C boiling point = -33.34 °C

Of the above, the compounds the one that closely resembles water is C₆H₆

Otrada [13]3 years ago

7 0

Answer:d

Explanation:

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Explanation:

A substance that is neither acidic nor basic is neutral. The pH scale measures how acidic or basic a substance is. The pH scale ranges from 0 to 14. A pH of 7 is neutral.

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3 years ago

A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its ve

Gemiola [76]

Answer:

(a) v = 5.42m/s

(b) vo = 4.64m/s

(d) Δy = 5.11*10^-3m

Explanation:

(a) The velocity of the ball before it hits the floor is given by:

$v=\sqrt{2gh}$ (1)

g: gravitational acceleration = 9.8m/s^2

h: height where the ball falls down = 1.50m

$v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}$

The speed of the ball is 5.42m/s

(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

$h_{max}=\frac{v_o^2}{2g}$ (2)

vo: velocity of the ball where it starts its motion upward

You solve for vo and replace the values of the parameters:

$v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}$

The velocity of the ball is 4.64m/s

(c) The acceleration is given by:

$a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}$

$a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}$

The acceleration of the ball is 2874.28/s^2

(d) The compression of the ball is:

$\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m$

THe compression of the ball when it strikes the floor is 5.11*10^-3m

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3 years ago

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Answer:

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