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Gekata [30.6K]
4 years ago
13

If a 100.0 g sample of water at 6.7°C is added to a 100.0 g sample of water at 57.0°C, determine the final temperature of the wa

ter. Assume no heat is lost to the surroundings.
Chemistry
2 answers:
just olya [345]4 years ago
5 0

Answer:

The answer to your question is  Te = 31.85 °C

Explanation:

Data

mass 1 = 100 g

temperature 1 = 6.7°C

mass 2 = 100 g

temperature 2 = 57°C

equilibrium temperature = ?

C = 1

Formula

                  mCΔT = - mCΔT

Substitution

                  100(Te - 6.7) = - 100(Te - 57)

Simplification

                  100 Te - 670 = - 100 Te + 5700

                  100Te + 100Te = 5700 + 670

                                 200Te = 6370

                                         Te = 6370 / 200

Result

                                          Te = 31.85 °C

Pavel [41]4 years ago
5 0

Answer:

The final temperature will be 31.85 °C

Explanation:

Step 1: Data given

Mass of water = 100.0 grams

Temperature of water = 6.7 °C

Mass of the other sample water = 100.0 grams

Temperature = 57.0 °C

Step 2: Calculate the final temperature

Energy Lost by the hot water = Energy Gain by the cold water

mass of hot water*c*(Temp. of hot water-x) = mass of cold water*c*(x-Temp. of cold water)

Energy Lost by the hot water=Energy Gain by the cold water

mass of hot water*c*(Temp. of hot water-x)=mass of cold water*c*(x-Temp. of cold water)

100*c*(57-x)=100*c*(x-6.7)

100*(57-x)=100*(x-6.7)

5700 - 100x = 100x -670

6370 = 200x

x = 31.85

The final temperature will be 31.85 °C

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A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate.

Solubility rules are used to predict whether some double-replacement reactions will occur.

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What is the percent composition of nitrogen in a 2.57 g sample of Al(NO3)3?
Lisa [10]

Answer:

19.8% of Nitrogen

Explanation:

In the Al(NO₃)₃ there are:

1 atom of Al

3 atoms of N

And 9 atoms of O

The molar mass of Al(NO₃)₃ is:

1 Al * (26.98g/mol) = 26.98g/mol

3 N * (14g/mol) = 42g/mol

9 O * (16g/mol) = 144g/mol

26.98 + 42 + 144 = 212.98g/mol

We can do a conversion using these molar masses to find the mass of nitrogen is the sample, that is:

2.57g * (42g/mol / 212.98g/mol) =

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Percent composition of nitrogen is:

0.51g N / 2.57g * 100

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6 0
3 years ago
An object has a mass of 6.8 g and volume of 34 mL. What is the density of the object?​
jenyasd209 [6]

Answer:

<h2>Density = 0.2 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

<h3>Density =  \frac{mass}{volume}</h3>

From the question the points are

mass = 6.8 g

volume = 34 mL

Substitute the values into the above formula and solve

That's

<h3>Density =  \frac{6.8}{34}</h3>

We have the final answer as

<h3>Density = 0.2 g/mL</h3>

Hope this helps you

7 0
3 years ago
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