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enot [183]
3 years ago
10

A laboratory experiment requires 250 millimeters of water for boiling. It also requires 100 millimeters of water for a cooling p

rocess.
Chemistry
1 answer:
Wewaii [24]3 years ago
7 0

Answer:

1.05 L

Explanation:

There is some info missing I think this is the original question.

<em>A laboratory experiment requires 250 millimeters of water boiling. it also requires 100 mills of water for a cooling process. If a student performs the experiment three times, how much total water will the student need? Give your answer in liters.</em>

<em />

Step 1: Calculate the volume required for each experiment.

The volume required is the sum of the volumes used: 250 mL + 100 mL = 350 mL

Step 2: Calculate the volume required for the 3 experiments

We have to multiply the volume required for each experiment by 3.

3 × 350 mL = 1050 mL

Step 3: Convert the volume to liters

We use the relation 1 L = 1000 mL.

1050 mL × (1 L/1000 mL) = 1.05 L

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Answer : The temperature of the chloroform will be, 101.67^oF

Explanation :

First we have to calculate the mass of chloroform.

\text{Mass of chloroform}=\text{Density of chloroform}\times \text{Volume of chloroform}=1.4832g/ml\times 74.81ml=110.958g

conversion used : (1cm^3=1ml)

Now we have to calculate the temperature of the chloroform.

Formula used :

q=m\times c\times (T_{final}-T_{initial})

where,

q = amount of heat or energy = 1.46 kJ = 1460 J   (1 kJ = 1000 J)

c = specific heat capacity = 0.96J/g.K

m = mass of substance = 110.958 g

T_{final} = final temperature = ?

T_{initial} = initial temperature = 25^oC=273+25=298K

Now put all the given values in the above formula, we get:

1460J=110.958g\times 0.96J/g.K\times (T_{final}-298)K

T_{final}=311.706K

Now we have to convert the temperature from Kelvin to Fahrenheit.

The conversion used for the temperature from Kelvin to Fahrenheit is:

^oC=\frac{5}{9}\times (^oF-32)

As we know that, K=^oC+273 or, K-273=^oC

K-273=\frac{5}{9}\times (^oF-32)

K=\frac{5}{9}\times (^oF-32)+273  ...........(1)

Now put the value of temperature of Kelvin in (1), we get:

311.706K=\frac{5}{9}\times (^oF-32)+273

T_{final}=101.67^oF

Therefore, the temperature of the chloroform will be, 101.67^oF

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