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3 years ago

How much water would I need to add to 500 mL of a 2.4 M KCl solution to make a 1.0 M solution?

Chemistry

1 answer:

GalinKa [24]3 years ago

3 0

Answer:

700 mL of water

Explanation:

This is the perfect example of dilation calculations. Along with this concept we have a formula c = n ( solute ) / V ( solution ). Let us first solve for n by changing this equation to isolate the solute,

n ( KCL ) = 2.4 mol / L * 500 * 10 ^ - 3 L,

n ( KCL ) = 1.2 moles ( KCL )

Knowing the amount of moles of potassium chloride, we have to now identify how much is present in the target solution,

V = 1 .2 moles / ( 1.0 moles / L )

V = 1.2 L = 1200 mL

_______________________________________________________

Vadded = 1200 - 500 = 700 mL

<u><em>Hope that helps!</em></u>

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Assuming you start with 10.0 g of zinc and 10.0 g hydrochloric acid, identify the limiting reagent and determine what mass of th

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The balanced chemical reaction for the substances given would be as follows:

Zn + 2HCl = ZnCl2 + H2

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3 years ago

How many moles are in 9.12 x 1023 molecules of sugar?

aleksandrvk [35]

Answer:

1.514 moles

Explanation:

For this problem you want to use dimensional analysis and cancel out your molecules of sugar and be left with moles of sugar. We know that 1 mole (of anything) = 6.022 x 10 ^ 23 molecules, so we should use that conversion to help us. Start with 9.12 x 10 ^23 molecules and divide by 6.022 x 10 ^ 23 molecules, and you will be left with moles.

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2 years ago

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3 years ago

2) A balloon was inflated to a volume of 5.0 liters at a temperature of

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Answer:

6.12 L

Explanation:

Given that,

Initial volume, V₁ = 5 L

Initial temperature, T₁ = 7.0°C = 343 K

Final temperature, T₂ = 147°C = 420 K

We need to find its new volume. The relation between volume and temperature is given by :

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{5\times 420}{343}\\\\V_2=6.12\ L$

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2 years ago