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Margaret [11]
4 years ago
11

Design an experiment to collect data that supports the claim that a 1.0M NaCl solution is a homogenous mixture.

Chemistry
1 answer:
Alika [10]4 years ago
7 0
The claim is that NaCl mixture is a homogeneous mixture.
Homogeneous mixture means that the components of the mixtures cannot be determined or separated by the naked eye. However, these components can be separated using physical means, such as boiling, evaporation and condensation which will be used in this experiment.

First, we need to prepare one molar solution of NaCl. To do so, we will dilute a mass of 58.44 grams (molar mass of NaCl) in 1 liter of water.
By this, we will have NaCl solution.

We can notice that once the NaCl is diluted in water, all what you can see is a clear solution. You cannot see the separate particles of NaCl in water.
..............> observation I

Now, we will heat this solution until it boils and water starts evaporating. We will place a cold surface above the steam coming out from the boiling solution.

What we will observe is that when all the water evaporates, we can see white precipitate of NaCl in the bottom of the container. Examining the cold surface placed above the steam, we can see that the water has condensed on this surface.
.........>observation II

Based on this, we managed to use boiling, evaporation and condensation (physical methods) to restore the components of the solution separately.
.............>conclusion

Based on observation I, observation II and the conclusion. we were able to prove that NaCl solution is a homogeneous mixture.
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3 0
4 years ago
What volume of a 0.00945-M solution of potassium hydroxicce would be required to titrae 50.00mL of a sample of acid rain with a
Zepler [3.9K]

<u>Answer:</u> The volume of HBr solution required is 130.16 mL

<u>Explanation:</u>

To calculate the concentration of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is KOH

We are given:

n_1=2\\M_1=123\times 10^{-4}M=0.0123M\\V_1=50mL\\n_2=1\\M_2=0.00945M\\V_2=?mL

Putting values in above equation, we get:

2\times 0.0123\times 50=1\times 0.00945\times V_2\\\\V_2=130.16mL

Hence, the volume of KOH solution required is 130.16 mL

5 0
4 years ago
How<br> many<br> molecules are in 8.55x10-15 moles of a substance
solniwko [45]

Answer:

Know this

1mole of a substance contains 6.022x10²³molecules

Now

8.55x10^-15moles x 6.022x10²³molecules/1mole

=5.15x10^9molecules.

6 0
4 years ago
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