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2 years ago

2. How many grams of NaCl are required to prepare 0.40 L of a 0.75 M solution?

1 answer:

4 0

The mass of a NaCl solution that is required to prepare 0.40 L of a 0.75 M solution is 17.55g. Details about mass can be found below.

<h3>How to calculate mass?</h3>

The mass of a substance can be calculated by multiplying the number of moles by its molar mass.

However, the number of moles of a solution must be initially calculated by using the following formula:

molarity = no of moles ÷ volume

no of moles = 0.75 × 0.40

no of moles = 0.3 moles

mass of NaCl = 0.3 × 58.5 = 17.55g

Therefore, the mass of a NaCl solution that is required to prepare 0.40 L of a 0.75 M solution is 17.55g.

Learn more about mass at: brainly.com/question/19694949

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GarryVolchara [31]

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Explanation:

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The total volume is given by:

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Then, the total volume of the Igloo is:

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Now, by using the density we can find the mass of the Igloo:

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2 years ago

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what i

Ostrovityanka [42]

Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :

[butane]=1.0 M , [isobutane]=2.5 M

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?

Answer:

The equilibrium concentration of each gas:

[Butane] = 1.14 M

[isobutane] = 2.86 M

Explanation:

Butane ⇄ Isobutane

At equilibrium

1.0 M 2.5 M

After addition of 0.50 M of butane:

(1.0 + 0.50) M -

After equilibrium reestablishes:

(1.50-x)M (2.5+x)

The equilibrium expression will wriiten as:

$K_c=\frac{[Isobutane]}{[Butane]}$

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