If 18.6 g of methanol is used to dissolve 2.68 g of #Hg(CN)_2#, what is the molality of the solution?
And this is where
The formula for "molality" is ((2.68g)/(252.65gmol^-1))/(18.6xx10^-3kg) =??mol*kg^-1?
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The molality of the solution is approximately 4.75 mol/kg.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
- What volume (in L) of a 3.95 M potassium chloride, #KCl#, solution would be needed to make 325 mL of a 2.76 M solution by dilution?
- How do you calculate the molarity of each solution 1.0 mol KCI in 750 mL of solution and then 0.50 mol #MgCl_2# in 1.5 L of solution?
- How many grams of #NaCl# should be weighed to prepare 1 L of 20 ppm solution of #Na^+#?
- How does a solution differ from a colloid?
- How do we do we demonstrate a #"saturated solution"#?

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