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3 years ago

A sample of water has a volume of 24.0 millimeters and a mass of 23.8 grams. what is the density

Chemistry

1 answer:

victus00 [196]3 years ago

3 0

24.0mm^3=24÷10÷10÷10cm^3
density=Mass÷Volume

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2055 Q. No. 10^-2

insens350 [35]

Answer:

Explanation:

Moles of KOH = $10^{-2}$

Volume of water = 10 liters

Concentration of KOH is given by

$[KOH]=\dfrac{10^{-2}}{10}\\\Rightarrow [KOH]=10^{-3}\ \text{M}$

$[KOH]$ is strong base so we have the following relation

$[KOH]=[OH^{-}]=10^{-3}\ \text{M}$

$pOH=-\log [OH^{-}]=-\log10^{-3}$

$\Rightarrow pH=14-3=11$

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Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points from highest to l

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Answer:

NH4Cl > Li2SO4 > CoCl3

Explanation:

Let us recall that the freezing point depression depends on the molality of the solution and the number of particles present.

Let us also recall that freezing point depression is a colligative property. It depends on the number of particles present in solution.

Usually, the more the number of particles present, the lower the freezing point. Hence, NH4Cl which has only two particles will have the highest freezing point while CoCl3 which has four particles will have the lowest freezing point.

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4 years ago

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Suppose 10.0 mL of 2.00 MNaOH is added to (a) 0.780 L of pure water and (b) 0.780 L of a buffer solution that is 0.682 Min butan

katrin2010 [14]

Answer:

a) pH will be 12.398

b) pH will be 4.82.

Explanation:

a) The moles of NaOH added = molarity X volume (L) = 2 X 0.01 = 0.02 moles

The total volume after addition of pure water = 0.780+0.01 = 0.79 L

The new concentration of /NaOH will be:

$molarity=\frac{molesofsolute}{volumeofsolution}=\frac{0.02}{0.79}=0.025M$

the [OH⁻] = 0.025

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b) The buffer has butanoic acid and butanoate ion.

i) Before addition of NaOH the pH will be calculated using Henderson Hassalbalch's equation:

$pH=pKa+log\frac{[salt]}{[acid]}$

pKa= $-logKa=-log(1.5X10^{-5})=4.82$

ii) on addition of base the pH will increase.

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4 years ago