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Savatey [412]
2 years ago
9

When the metal was placed in the calorimeter its

Chemistry
2 answers:
liq [111]2 years ago
7 0

Answer:

-79.6C    

-80C  

-81.2C

Explanation:

guajiro [1.7K]2 years ago
5 0

Answer:

Exam 3 Material

Homework Page Without Visible Answers

This page has all of the required homework for the material covered in the third exam of the first semester of General Chemistry. The textbook associated with this homework is CHEMISTRY The Central Science by Brown, LeMay, et.al. The last edition I required students to buy was the 12th edition (CHEMISTRY The Central Science, 12th ed. by Brown, LeMay, Bursten, Murphy and Woodward), but any edition of this text will do for this course.

Note: You are expected to go to the end of chapter problems in your textbook, find similar questions, and work out those problems as well. This is just the required list of problems for quiz purposes. You should also study the Exercises within the chapters. The exercises are worked out examples of the questions at the back of the chapter. The study guide also has worked out examples.

These are bare-bones questions. The textbook questions will have additional information that may be useful and that connects the problems to real life applications, many of them in biology.

Explanation:

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A mixture of gases contains 4.46 moles of neon (Ne), 0.74 mole of argon (Ar), and 2.15 moles of xenon (Xe). Calculate the partia
viva [34]

Explanation:

The partial pressure of an individual gas is equal to the total pressure of the mixture multiplied by the mole fraction of the gas.

Total pressure = 2atm

Mole Fraction = number of moles / total number of moles

Neon

Mole Fraction = 4.46 / 7.35 = 0.607

Partial Pressure = 0.607 * 2 = 1.214 atm

Argon

Mole Fraction = 0.74 / 7.35 = 0.101

Partial Pressure = 0.101 * 2 = 0.202 atm

Xenon

Mole Fraction = 2.15 / 7.35 = 0.293

Partial Pressure = 0.293 * 2 = 0.586 atm

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3 years ago
For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts.
Diano4ka-milaya [45]
The balanced chemical reaction is given as follows:

<span>2 KClO3(s) → 2 KCl(s) + 3 O2(g)

The starting amount of the reactant are given above. These values would be used for the calculations. We do as follows:

</span>2.72 g KClO3 (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 1.06 g O2
<span>
0.361 g KClO3 </span>(1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 0.14 g O2
<span>
83.6 kg KClO3 (1000g / 1kg) </span>(1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 3275.76 g O2
<span>
22.5 mg KClO3</span> (1 g / 1000 mg) (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 0.009 g O2
4 0
2 years ago
Read 2 more answers
The surface ocean currents move heat energy away from where?
Dmitriy789 [7]
Away from the equator.
8 0
3 years ago
If the oxidation number of nitrogen in a certain molecule changes from +3 to -2 during a reaction, is the nitrogen oxidized or r
aliina [53]
The nitrogen has been reduced or has undergone reduction and it has gained one electron
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2 years ago
Why is it better to measure the percent change in mass<br> ratherthan just using the change in mass?
never [62]

Answer: The result is presented in proportion which gives a clearer understanding and accurate result.

Explanation: Percentage change in mass is the proportion of the initial mass of a substance changed after sometime. The results is presented as a percentage making it more accurate and can help to give future reference to weight calculations.

Change is Mass is the mass of a substance left after sometime mostly given in grams. It is not as accurate as percentage change in mass. It is generally better to show results in percentage change in mass as it gives a better understanding of what mass of a substance was lost after a given period or after application of energy like Heat or increased temperature.

6 0
3 years ago
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