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Mashutka [201]
3 years ago
7

A pair of students determining the molarity of their unknown HCl solution calculates the concentration to be 0.0961 M on their f

irst trial and 0.104 M on their second trial. Do they need to run a third trial
Chemistry
2 answers:
Varvara68 [4.7K]3 years ago
7 0

Answer:

Yes.

Explanation:

It should be noted that the meaning of molarity is the ratio of moles of solute per liter of solution.

It should be understood that when determining or finding the molarity of an unknown compound ,the process should be performed or carried out at least 3 times. This is done to remove any form of doubt.

The first calculated value for the concentration of the compound will be regarded as rough value, while the second and the third will be regarded as the first and second values respectively.

In this case, the third value for the concentration of HCl will be calculated to for confirmation of other value, that is to be finally sure of its concentration.  

Katyanochek1 [597]3 years ago
5 0

Answer:

Yes.

Explanation:

For a typical experiment, you should plan to repeat it at least three times (more is better).

The value gotten in the first trial is not close to the value gotten from the second trial.

The solution to these problems is to do repeated trials. Repeated trials are when you do a measurement multiple times - at least three, commonly five, but the more the better. When you measure something once, the chance that the number you get is accurate is much lower. But if you measure it several times, you can take an average of those numbers and get a result that is much closer to the truth.

Repeating a trial gives a RELIABLE result.

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2 years ago
What volume of 3.00 M HCl will form a solution with an acidic pH when mixed with 100 mL of 3.00 M NaOH
Maksim231197 [3]

<u>Answer:</u> The volume of acid should be less than 100 mL for a solution to have acidic pH

<u>Explanation:</u>

To calculate the volume of acid needed to neutralize, 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 HCl

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

We are given:

n_1=1\\M_1=3.00M\\V_1=?mL\\n_2=1\\M_2=3.00M\\V_2=100mL

Putting values in above equation, we get:

1\times 3.00\times V_1=1\times 3.00\times 100\\\\V_1=\frac{1\times 3.00\times 100}{1\times 3.00}=100mL

For a solution to be acidic in nature, the pH should be less than the volume of acid needed to neutralize.

Hence, the volume of acid should be less than 100 mL for a solution to have acidic pH

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