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Reptile [31]
3 years ago
12

What is the name of this molecule?

Chemistry
2 answers:
ankoles [38]3 years ago
7 0

Answer: The answer is either A or C. I'm leaning more towards A.

Svet_ta [14]3 years ago
5 0

Answer:

I think its C :)

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What is the final concentration when 5 ml of a 2.5M copper sulphate solution is diluted to 750 ml?
Firdavs [7]

Answer: The final concentration when 5 ml of a 2.5M copper sulphate solution is diluted to 750 ml is 0.017 M

Explanation:

According to the dilution law,

M_1V_1=M_2V_2

where,  

M_1 = molarity of stock CuSO_4 solution = 2.5 M

V_1 = volume of stock CuSO_4solution = 5 ml  

M_1 = molarity of diluted CuSO_4 solution = ?

V_1 = volume of diluted CuSO_4 solution = 750 ml

Putting in the values we get:

2.5\times 5=M_2\times 750

M_2=0.017

Therefore the final concentration when 5 ml of a 2.5M copper sulphate solution is diluted to 750 ml is 0.017 M

6 0
3 years ago
What is known as the amount of<br> mass in a given volume?
Ganezh [65]

Answer:

Density

Explanation:

Density is a measure of how much mass is contained in a given volume

3 0
3 years ago
Sponges, Cnidarians, Flatworms, and Roundworms are all in which group of animals?
Westkost [7]
These animals are all invertebrates
3 0
3 years ago
Tia has a sample of pure gold (Au). She weighed the sample and the result was 62.4 grams. Tia wants to determine the number of a
Marta_Voda [28]
I hope you are able to find your answer through the guidance of this site as I have yet to cover this topic in chemistry: http://scientifictutor.org/1021/chem-how-to-convert-between-grams-and-molecules/
7 0
3 years ago
What molarity of nitric acid (HNO3) was used if 2.00 L must be used to prepare 4.5 L of a 0.25 M HNO3 solution?
Ymorist [56]

The   molarity  of (HNO₃) that was used  if 2.00 L must  be used   to prepare 4.5 L  of a 0.25M HNO₃ solution is   0.563 M


 <u><em>calculation</em></u>

  This is calculated  usind  M₁V₁=M₂V₂  formula

where,

         M₁(  molarity ₁) = ?

         V₁( volume ₁) = 2.00 L

        M₁ (molarity ₂) = 0.25M

         V₂( volume₂) = 4.5 L

make M₁ the subject  of the formula by  diving both side of the formula  by V₁

   M₁  is therefore = M₂V₂/V₁

M₁ =[ (0.25 M  x 4.5 L) / 2.00 L ]  =0.563 M

5 0
3 years ago
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