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Mazyrski [523]
3 years ago
10

A teacher wants to figure out the amount of a solution that is needed for a titration experiment in a class lab. The class has 1

5 students, and each experiment needs 50 mL of solution.
If each student will perform the experiment two times, how many mL of solution will the teacher need for the whole class?
Chemistry
2 answers:
LUCKY_DIMON [66]3 years ago
7 0

The correct answer is C. 1500 mL on e2020.

zmey [24]3 years ago
6 0
You need to multiply 15 by 2 since each student is doing 2 trials which gives you 30 (you can think of it as 30 students doing it once if that helps you).  Then you have to multiply 30 by 50 since the experiment is being done 30 times and each time requires 50mL which equals 1500mL.
Therefore the teacher needs to make 1500mL of solution for the lab.

I hope this helps.
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A sample of nitrogen gas is at a temperature of 50 c and a pressure of 2 atm. If the volume of the sample remains constant and t
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Answer:

The new temperature of the nitrogen gas is 516.8 K or 243.8 C.

Explanation:

Gay-Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of collisions with the walls increases, that is, the pressure increases. That is, the pressure of the gas is directly proportional to its temperature.

Gay-Lussac's law can be expressed mathematically as follows:

\frac{P}{T} =k

Where P = pressure, T = temperature, K = Constant

You want to study two different states, an initial state and a final state. You have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment. By varying the temperature to a new value T2, then the pressure will change to P2, and the following will be fulfilled:

\frac{P1}{T1} =\frac{P2}{T2}

In this case:

  • P1= 2 atm
  • T1= 50 C= 323 K (being 0 C= 273 K)
  • P2= 3.2 atm
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Replacing:

\frac{2 atm}{323 K} =\frac{3.2 atm}{T2}

Solving:

T2*\frac{2 atm}{323 K} =3.2 atm

T2=3.2 atm*\frac{323 K}{2 atm}

T2= 516.8 K= 243.8 C

<u><em>The new temperature of the nitrogen gas is 516.8 K or 243.8 C.</em></u>

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