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mash [69]
3 years ago
12

how many moles of cuo can be produced from 0.900 moles of Cu2O in the following reaction: 3Cu2O + O2--> 4 CuO

Chemistry
1 answer:
Tanya [424]3 years ago
5 0
Answer:
             1.8 mol of CuO

Solution:

The balanced chemical equation is as follow,

                               2 Cu₂O  +  O₂     →     4 CuO

According to equation,

                     2 moles of Cu₂O produced  =  4 moles of CuO
So,
          0.900 moles of Cu₂O will produce  =  X moles of CuO

Solving for X,
                     X  =  (0.900 mol × 4 mol) ÷ 2 mol

                     X  =  1.8 mol of CuO
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3 years ago
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Katyanochek1 [597]

Answer:

The answer to your question is     V2 = 4.97 l

Explanation:

Data

Volume 1 = V1 = 4.40 L                    Volume 2 =

Temperature 1 = T1 = 19°C               Temperature 2 = T2 = 37°C

Pressure 1 = P1 = 783 mmHg           Pressure 2 = 735 mmHg

Process

1.- Convert temperature to °K

T1 = 19 + 273 = 292°K

T2 = 37 + 273 = 310°K

2.- Use the combined gas law to solve this problem

                  P1V1/T1  = P2V2/T2

-Solve for V2

                  V2 = P1V1T2 / T1P2

-Substitution

                  V2 = (783 x 4.40 x 310) / (292 x 735)

-Simplification

                 V2 = 1068012 / 214620

-Result

                 V2 = 4.97 l

6 0
3 years ago
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Bogdan [553]
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3 years ago
Which would be the best to neutralize a large acid spill in your school lab: sodium hydroxide or baking soda? Explain.
nadya68 [22]

Consider the acid spill. It is already starting to do nasty things to, say, the floor or counter. So you grab the bottle of 10% NaOH and pour some on the spill. All of a sudden, you get a great deal of heat, and you don't have any visual evidence whether your put on too little or too much. But you have added more liquid to the spill, generated more heat, and will get more damage. You have made a bigger mess, and if you added too much, you then have a neutralization problem to deal with.  

And if it is something like a strong sulfuric acid solution, adding sodium hydroxide solution will be extremely exothermic, and you could get some really nasty results.  

So now approach the spill with a handful of baking soda. You sprinkle it on the spill. It fizzes, and carbon dioxide is given off. That actually, in a very tiny way, moderates the temperature of the neutralization. And you can keep adding baking soda until the fizzing stops, and then perhaps some water to mix everything well. But what you have done is kept the volume to a minimum, added a neutralization agent that has a visible endpoint (no more gas being given off), and you don't suddenly have a huge amount of highly basic solution because you added too much.  

And what is also nice about baking soda is that you can toss some with your hand or even with a spoon, and get some distance from the spill. With a liquid, you have to get much closer

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5 0
3 years ago
General Chemistry fourth edition by McQuarrie, Rock, and Gallogly. University Science Books presented by Macmillan Learning.
Helen [10]

Answer:

3.07 Cal/g

Explanation:

Step 1: Calculate the heat absorbed by the calorimeter

We will use the following expression.

Q = C × ΔT

where,

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Q = 37.60 kJ/°C × 2.29 °C = 86.1 kJ

According to the law of conservation of energy, the heat released by the candy has the same magnitude as the heat absorbed by the calorimeter.

Step 2: Convert 86.1 kJ to Cal

We will use the conversion factor 1 Cal = 4.186 kJ.

86.1 kJ × 1 Cal/4.186 kJ = 20.6 Cal

Step 3: Calculate the number of Cal per gram of candy

20.6 Cal/6.70 g = 3.07 Cal/g

3 0
3 years ago
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