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Salsk061 [2.6K]
3 years ago
10

PLEASE HELP BEFORE 7 A.M. PACIFIC TIME SEE ATTACHED.

Chemistry
2 answers:
pickupchik [31]3 years ago
7 0

Answer:

(i) 32.9 mL; (ii)  37.5 mL; (iii) 21.73 mL

Explanation:

You should always try to read a measuring instrument to a tenth of the smallest scale division.

Here, you are measuring liquids, so you take the scale reading from the bottom of the meniscus.

(i) Graduated cylinder

There are 10 divisions between 30 mL and 40 mL, so each division represents 1 mL.

The level of the liquid appears to be between 32 mL and 33 mL. It is much closer to 33 mL (perhaps right on 33 mL).

You should report the volume to the nearest 0.1 mL. I would read the volume as 32.9 mL, but 32.8 and 33.0 are also acceptable.

Note: If you think the level is right on the 33 mark, you report the volume as 33.0 mL (NOT 33 mL).

(ii) Thermometer

The reading is about half-way between 87 ° and 88 °.

I would report the temperature as 87.5 °, but 87.4 ° and 87.6 ° would also be acceptable.

(iii) Buret

There are 10 divisions between 21 mL and 22 mL, so each division represents 0.1 mL.

You should estimate to the nearest 0.01 mL.

The liquid level is about a third of the way from 21.7 mL to 21.8 mL.

I would report the volume as 21.73 mL, but 21.72 mL and 21.74 mL are also acceptable.

 

anastassius [24]3 years ago
5 0
For i: 33mL
For ii: 87-88mL
For iii:22.3mL
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The mass of water produced is 792 grams by the combustion of 568 grams of decane.

Given:

Combustion of 568 grams of decane with 2979 grams of oxygen.

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To find:

The mass of water produced by combustion of 568 grams of decane.

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According to reaction, 2 moles of decane reacts with 31 moles of oxygen, then 4 moles of decane will react with:

=\frac{31}{2}\times 4mol=62\text{ mol of}O_2

But according to the question, we have 93.0 moles of oxygen gas which is more than 62 moles of oxygen gas.

So, this means that oxygen gas is present in an excessive amount. Which simply means:

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According to reaction, 22 moles of water is produced from 2 moles of decane, then 4 moles of decane will produce:

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792 grams of water is produced by the combustion of 568 grams of decane.

Learn more about limiting reagent and excessive reagent here:

brainly.com/question/14225536?referrer=searchResults

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