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Lesechka [4]
3 years ago
15

A student assistant is cleaning up after a chemistry laboratory exercise and finds three one-liter bottles containing alcohol so

lutions. The first bottle is one-half full with a 20% alcohol solution, the second bottle is one-fifth full with a 30% alcohol solution, and the third bottle is one-tenth full with a 50% alcohol solution. The student pours the contents of all three bottles into an empty one-liter bottle. What is the approximate alcohol content of this mixed solution
Chemistry
1 answer:
Svetllana [295]3 years ago
3 0

The approximate alcohol content is 210 ml.

Explanation:

It can be deduced from the question that each bottle is of 1000ml or 1 litre.

The first bottle is one half full means it has 500 ml of solution and it has 20% alcohol in it. So volume of alcohol in the solution is

20/100*500

=100 ml

The first bottle is one fifth full, so the volume of mixture is 1/5th of 1000ml

so it is 200ml having 30% alcohol

30/100*200

= 60 ml

The third bottle is one tenth full so its volume is 1/10*1000

100 ml.  having 50% of alcohol

50/100*100

50 ml.

The alcohol content obtained from all these 3 litres is:

100+60+50

= 210 ml of alchohol is obtained from 800 ml of mixture.

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Answer: Na, S, Cl

Explanation:

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Sodium (Na) comes before Sulfur (S) which comes before Chlorine (S) so this is the decreasing order as they are all in the same period.

7 0
2 years ago
What is the molality of a solution that contains 1.34 moles of NaCl in 2.47 kg of solvent
dsp73

The molality is 0.54 M when 1.34 moles of NaCl is present in 2.47 kg of solvent.

<u>Explanation:</u>

Molality is the measure of how much of amount of solute is dissolved in the solvent. So it is calculated as the ratio of moles of solute to the grams of solvent.

         \text {Molality}=\frac{\text {Moles of solute}}{\text {Mass of solvent}}

As in this case, the solute is NaCl and solvent is unknown. So the moles of solute is given as 1.34 moles and the mass of solvent is given as 2.47 kg.

Hence, \text { molality }=\frac{1.34}{2.47}= 0.54 \mathrm{M}

Thus, the molality is 0.54 M when 1.34 moles of NaCl is present in 2.47 kg of solvent.

8 0
3 years ago
Determine the number of significant figures in the measurement quantity: 2600. kg.
xxMikexx [17]

Answer:

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Explanation:

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For example, 1.8 → 2 significant figures

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For example, 3.02 → 3 significant figures

When you have 0 on the left in the measure, we do not consider as a sgnificant figures

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3.0 → 2 significant figures (0 is on the right, not the left)

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3 years ago
What are three areas are the natural sciences commonly divided
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3 years ago
Write this number in standard notation.<br> 1.986 x 10negative exponent 6 or 1.986 x 10^6
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Answer:

0.000001986

Explanation:

We have the number 1.986*10^{-6}. Standard notation would basically be this number without the 10^{-6} part.

To get rid of this part, we need to move the decimal point 6 places to the left (we go left because it's negative 6, indicating division).

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