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Verdich [7]
3 years ago
5

How does an acid–base indicator react when placed in an acidic solution?

Chemistry
2 answers:
Anestetic [448]3 years ago
4 0
An indicator changes color when placed in an acidic solution
Minchanka [31]3 years ago
3 0

Answer: Option (a) is the correct answer.

Explanation:

An acid-base indicator is an indicator which changes color according to the pH of solution in which it is placed.

For example, phenolphthalein is an acid-base indicator.

Thus, we can conclude that out of the given options, an acid–base indicator react by changing color when placed in an acidic solution, is the correct option.

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How to atoms form a new substance
marin [14]

Answer:

they join together

Explanation:

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Where does energy in an ecosystem come from?
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Energy in an ecosystem comes from the Sun.

Explanation:

It is a major source of energy for organisms, plants and the whole ecosystem basically.

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A student mixed some yellow sulfur powder with some iron filings. She was able to take the iron out of the sulfur by using a mag
Kay [80]

Answer: Chemical change

Explanation:

Physical change is a change in which there is no rearrangement of atoms and thus no new substance is formed. There is only change in physical state of the substance. The change can be reversed using physical methods.

Chemical change is a change in which there is rearrangement of atoms and thus new substance is formed. There may or may not be a change in physical state. The change can only be reversed using chemical methods.

Thus when iron and sulfur are heated in a test tube, the chemical change occurs and thus lead to formation of new substances which could not be separated using magnet.

8 0
2 years ago
What part of the sun do we see from earth?
NISA [10]

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photosphere

Explanation:

photosphere

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8 0
3 years ago
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Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h
bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

We know that,

The relation between the C_p\text{ and }C_v for an ideal gas are :

C_p-C_v=R

As we are given :

C_p=28.253J/K.mole

28.253J/K.mole-C_v=8.314J/K.mole

C_v=19.939J/K.mole

Now we have to calculate the entropy change of the gas.

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. (w=-pdV)

(C) Heat during the process will be,

q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

7 0
3 years ago
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