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creativ13 [48]
3 years ago
10

What volume in mL of a 2.9M solution is needed in order to prepare 2.3L of a 1.1M solution?

Chemistry
2 answers:
Mumz [18]3 years ago
7 0
V - 0.87 liters I'm sure I'm right 
sladkih [1.3K]3 years ago
6 0
As i am reading the question, I notice they give two concentrations (M), one volume (L), and they are asking for another volume. this is a hint that you need to use the following formula

C1V1 = C2V2                (note: C stands for concentration)

C1= 2.9 M
V1= ?
C2= 1.1 M
V2= 2.3 L

now let's plug in the numbers

( 2.9 x V1)= (1.1 X 2.3)

V1= 0.87 Liters


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The mass of 2.15 mol of hydrogen sulphide (H₂S) will be 73.272 gm and the mass of  3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) will be 1.82 gm

<h3>What is Mole ?</h3>

A mole is a very important unit of measurement that chemists use.

A mole of something means you have 6.023 x 10 ²³ of that thing.

  • For 2.15 mol of hydrogen sulphide (H₂S) :

1 mole hydrogen sulphide (H₂S) = 34.08088 grams

Therefore,

2.15 mol of hydrogen sulphide (H₂S) = 34.08088 grams x 2.15 mol

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1 mol of lead(II) iodide, (PbI₂) = 461.00894 grams

Therefore,

3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) = 461.00894 grams x 3.95 × 10⁻³ mol

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Hence,The mass of 2.15 mol of hydrogen sulphide (H₂S) will be 73.272 gm and the mass of  3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) will be 1.82 gm

Learn more about mole here ;

brainly.com/question/21323029

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