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iris [78.8K]
3 years ago
5

What are 2 ways to give an object potential energy?

Chemistry
1 answer:
natita [175]3 years ago
7 0
There are several ways to give an object potential energy. One can move the object against the force of gravity to increase. One can also stretch an object out or put pressure on it.
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Match each organ to its function. Tiles bone heart stomach lung Pairs pumps blood through the body Both provides structure for t
solniwko [45]
Heart: Pumps blood through the body
Bone: Provides structure
Stomach: Breaks down food into small particles
Lungs: Oxygenates blood
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3 years ago
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Why is salt added to hot chicken soups during cooking besides giving the chicken soups a good taste?
SIZIF [17.4K]

Answer:

The answer is option 3.

Explanation:

When salt is added to the water, the boiling point increases because it needs to take in more energy from heat to <u>b</u><u>r</u><u>e</u><u>a</u><u>k</u><u> </u><u>d</u><u>o</u><u>w</u><u>n</u> the bonds and dissolve the salt in the water.

(Correct me if I am wrong)

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2 years ago
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How dose heat travel between objects that are touching one another?
11Alexandr11 [23.1K]
Through the process of conduction, heat van be trnafered between two solids.
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3 years ago
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Free Energy
ratelena [41]

Answer:

a) galvanic cell

b)electrolytic cell

c) i) K=6.27x10'34

ΔG°=198790 J

ii) K=3.58x10'-34

ΔG°= 191070 J

d) E°=0.278 v

ΔG°= -26827 J

Explanation:

a) There are two kinds of an electrochemical cell, the first is called "galvanic cells", and the second "electrolytic cell".

The fuel cells are capable of produce electric energy through chemical reactions. These reactions are often spontaneous. So, the galvanic cell has a negative value for Gibbs free energy.

b) The electrolytic cell increases the value of Gibbs energy, to positive values, due to the reactions are not spontaneous.

c) i) look image attached

ii) k = look image attached

ΔG° = -nFE° = - 6 X 95500 J/vmole x (-0.33 v)

ΔG° =-191070

d) E°= 0.0592 v/n x lg K

E°= 0.0592V / 1 X log 5.0X10'4

E°= 0.278 v

ΔG° = -nFE° = -1 x 96500 J/ vmole x 0.278v

ΔG° = -26827 J

5 0
3 years ago
g If a small amount of a strong acid is added to a buffer made up of a weak acid, HA, and the sodium salt of its conjugate base,
ikadub [295]

Answer:

Explanation is in the answer

Explanation:

The pH of the buffer solution does not change appreciably because the strong acid (free H⁺) reacts with conjugate base of buffer producing more weak acid. pH formula of buffers is (Henderson-Hasselbalch formula):

pH = pKa + log ( [A⁻] / [HA] )

The addition of strong acid decreases [A⁻] increasing [HA]. pH change just in the log of the ratio of [A⁻] with [HA], that is a real little effect over pH of the buffer solution.

5 0
3 years ago
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