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sp2606 [1]
3 years ago
6

A researcher dilutes 30.0 ml of ethanol to 300.0 ml with distilled water. what is the percentage concentration by volume of the

resulting solution?9.09.10%
Chemistry
2 answers:
algol [13]3 years ago
4 0

Answer:

B. 10.0%

Explanation:

This is the answer on edg

Len [333]3 years ago
3 0

Answer: 9.09 %

Explanation:

To calculate the  percentage concentration by volume, we use the formula:

\text{Volume percent of solution}=\frac{\text{Volume of solute}}{\text{Volume of solution}}\times 100

Volume of ethanol (solute) = 30 ml

Volume of water (solvent) = 300 ml

Volume of solution= volume of solute + volume of solution = 30+ 300 = 330 ml

Putting values in above equation, we get:

\text{Volume percent of solution}=\frac{30}{30+300}\times 100=9.09\%

Hence, the volume percent of solution will be 9.09 %.

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How many molecules are in 2.38g of SO2
Murrr4er [49]

2.63 x 10^22 molecules of SO2.

To find this, start with what you know.

2.38g of SO2.

You need to first convert this into Moles since you cannot directly convert grams into molecules. In order to convert grams to moles, you need to find the molecular mass of SO2 - 64.066.

This is because Sulfur has the mass of 32.066 and Oxygen has a mass of 16, but since there are two Oxygen atoms, it's going to be 32.

Your equation should currently appear as so:

2.38g of SO2 = 1 Mole of SO2 / 64.066

Now, you need to convert this to molecules.

Whenever you are searching for molecules without a given amount, you always use Avogadro's number: 6.02 x 10^23

Now, your equation should appear as so:

2.38g SO2 = 1 Mole of SO2 / 64.066 next to a new fraction which is 6.02 x 10^23 / 1 Mole of SO2

Now, multiply across (2.38 x 1 x 6.02 x 10^23). When using Avogadro's number, don't forget to place parenthesis around it.

Then, divide that number by the bottom: 64.066.

Thus, your final answer is that there is 2.63 x 10^22 molecules of SO2.

Don't forget your units!

Hope this helps!

5 0
3 years ago
If there is currently 50kg of U-235 present in Oklo, how much must have been present 750 million years ago when the reaction too
Viefleur [7K]
To answer this question, you need to know the concept of half-life, which is how a radioactive material decreases in mass over time.

The half life of U-235 is 703.8 million years. The first part of this problem is to find the scale factor. To do this, divide the time that has past by the half life, like this:

\frac{750}{703.8}  = 1.066
Now, take this scale factor and multiply it by the current mass, like this:

50 \times 1.066 = 53.3
This number is what you add to the current mass to get the original mass. That is because the scale factor showed us that it was just over one half life. Since after one half life, the mass is cut in half, and this is over one half life, when we add to the original it will be a little over double. This equation illustrates the final addition:

50 + 53.3 = 103.3
I hope this helped you. Fell free to ask any further questions.
7 0
3 years ago
What are the 3 primary consumers and what are the 3 secondary consumers
goldenfox [79]
Primary:
Grasshopper
Mouse
Grass

Secondary:
Hawk
Snake
Coyote 
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3 years ago
Which type of salt would be most effective for melting ice on roads?
OLga [1]
Calcium Chloride is the salt that is effective for melting ice on roads

3 0
4 years ago
Read 2 more answers
Help with this please.
chubhunter [2.5K]

Answer:

a

Explanation:

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