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sergejj [24]
3 years ago
15

Some fertilizer blends contain magnesium nitrate (MgNO3). Suppose that a chemist has 1.24 liters of a 2.13 M solution of magnesi

um nitrate. If the chemist dilutes the solution to 1.60 M, what is the volume of the new solution?
Chemistry
2 answers:
tamaranim1 [39]3 years ago
7 0

Initial concentration of magnesium nitrate M1 = 2.13 M

Initial volume of magnesium nitrate, MgNO3  V1 = 1.24 L

Final concentration of MgNO3, M2 = 1.60 M

Let the final volume of MgNO3 upon dilution be V2

Formula to use:

M1*V1 = M2*V2

V2 = M1*V1/M2

      = 2.13 M * 1.24 L/1.60 M = 1.65 L

Thus, the final volume of magnesium nitrate solution upon dilution is 1.65 L

Eduardwww [97]3 years ago
6 0

Initial concentration of magnesium nitrate M1 = 2.13 M

Initial volume of magnesium nitrate, MgNO3  V1 = 1.24 L

Final concentration of MgNO3, M2 = 1.60 M

Let the final volume of MgNO3 upon dilution be V2

Formula to use:

M1*V1 = M2*V2

V2 = M1*V1/M2

      = 2.13 M * 1.24 L/1.60 M = <em>1.65 L</em>

<em>I believe I did it right...</em>

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Answer:

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Explanation:

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What volume will 1.27 moles of helium gas occupy at 80.00 °C and 1.00 atm?
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36.8 L

Explanation:

We'll begin by converting 80 °C to Kelvin temperature. This can be obtained as follow:

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Finally, we shall determine the volume occupied by the helium gas. This can be obtained as follow:

Number of mole (n) = 1.27 moles

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DIRECTIONS: Solve the following problems by using G.R.E.S.A (Given, Required,Equation, Solution, and Answer).
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Answer:

SEE BELOW

Explanation:

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6 0
2 years ago
Read 2 more answers
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