3 years ago

90 grams of NaNO3 were added to solution at 10 degrees Celsius. What type of solution was formed?

Chemistry

1 answer:

vichka [17]3 years ago

4 0

Answer:

friend me on here and imma send you the link Explanation:

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How many moles of KNO3 are needed to make 600 ml of a 1.3M solution?

ludmilkaskok [199]

M = n / V

Where, M is molarity (M or mol/L), n is number of moles of the solute (mol) and V is volume of the solution (L).

Here the solute is KNO₃.
The given molarity is 1.3 M
This means 1L of solution has 1.3 moles of KNO₃.

Hence moles in 600 mL = 1.3 M x 0.6 L = 0.78 mol

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When a 10 mL graduated cylinder is filled to the 10 mL mark, the mass of the water was measured to be 9.925 g. If the density of

Vlada [557]

Measured volume = 10 ml
mass = 9.925 g
density = 0.9975 g/ml
density = $\frac{mass}{volume}$
so actual volume = $\frac{mass}{density}$ = $\frac{9.925}{0.9975}$ = 9.95 mL
Percentage Error = $\frac{measured volume - Actual volume}{actual volume}$ x 100
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