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Elodia [21]
2 years ago
6

if an unknown sample solution is prepared by diluting 10.00 ml of the original solution to a total volume of 100.0 ml with deion

ized water, what is the dilution factor?
Chemistry
1 answer:
Usimov [2.4K]2 years ago
6 0

The dilution factor of the unknown sample is 10. The dilution factor of a solution refers to the ratio of the final volume of the now diluted solution to the initial volume of the of the initial concentrated solution.

Mathematically;

The dilution factor is given by the formula;

Dilution factor = Final volume of the now diluted solution/ Initial volume of more concentrated solution

Final volume of the now diluted solution = 100.0 ml

Initial volume of more concentrated solution = 10.00 ml

Dilution factor = 100.0 ml/10.00 ml

Dilution factor =  10

Learn more: brainly.com/question/20113402

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A chemist adds 0.60L of a 0.20/molL sodium thiosulfate Na2S2O3 solution to a reaction flask. Calculate the millimoles of sodium
Serggg [28]

Answer:

1.2×10² mmole of Na₂S₂O₃

Explanation:

From the question given above, the following data were obtained:

Volume = 0.6 L

Molarity = 0.2 mol/L

Mole of Na₂S₂O₃ =?

Molarity is simply defined as the mole of solute per unit litre of water. Mathematically, it is expressed as:

Molarity = mole /Volume

With the above formula, we can obtain the number of mole of Na₂S₂O₃ in the solution as illustrated below:

Volume = 0.6 L

Molarity = 0.2 mol/L

Mole of Na₂S₂O₃ =?

Molarity = mole /Volume

0.2 = Mole of Na₂S₂O₃ / 0.6

Cross multiply

Mole of Na₂S₂O₃ = 0.2 × 0.6

Mole of Na₂S₂O₃ = 0.12 mole

Finally, we shall convert 0.12 mole to millimole (mmol). This can be obtained as follow:

1 mole = 1000 mmol

Therefore,

0.12 mole = 0.12 mole × 1000 mmol / 1 mole

0.12 mole = 120 = 1.2×10² mmole

Thus, the chemist added 1.2×10² mmole of Na₂S₂O₃

7 0
3 years ago
According to the kinetic-molecular theory, how do gas particles behave?
mezya [45]

Answer:

They are in constant motion.

Explanation:

More energy\heat= more kinetic energy=more motion\movement

7 0
3 years ago
..can someone answer these questions ive already posted this question 16 times and I dont have much points left so please anyone
noname [10]

Answer

  1. A
  2. D
  3. D
  4. A
  5. B
  6. A

This is what I got, but i'm mot sure if I'm right

7 0
3 years ago
Read 2 more answers
How many milliliters of a 5.0 M H2SO4 stock solution would you need to prepare 108.0 mL of 0.45 M H2SO4?
PIT_PIT [208]
For the purpose we will use solution dilution equation:
c1xV1=c2xV2
Where, c1 - concentration of stock solution; V1 - a volume of stock solution needed to make the new solution; c2 - final concentration of new solution; V2 - final volume of new solution.
c1 = 5.00 M
c2 = 0.45 M
V1 = ?
V2 = 108 L
When we plug values into the equation, we get following:
5 x V1 = 0.45 x 108
<span>V1 = </span>9.72 L
7 0
3 years ago
The blank solution used to calibrate the spectrophotometer is 10.0 mL of 0.2 M Fe(NO3)3 diluted to 25.0 mL with 0.1 M HNO3. Why
Sliva [168]
<span>FeNCS+ product...............thats how you do it i believe </span>
4 0
2 years ago
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