Question

1. A 14.9 g sample of an aqueous solution of hydrobromic acid contains an unknown amount...

1.
A 14.9 g sample of an aqueous solution of hydrobromic acid contains an unknown amount of the acid.
If 13.5 mL of 0.583 M sodium hydroxide are required to neutralize the hydrobromic acid, what is the percent by mass of hydrobromic acid in the mixture?

2.A 14.0 g sample of an aqueous solution of nitric acid contains an unknown amount of the acid.
If 26.6 mL of 0.725 M barium hydroxide are required to neutralize the nitric acid, what is the percent by mass of nitric acid in the mixture?

3.An aqueous solution of barium hydroxide is standardized by titration with a 0.161 M solution of perchloric acid.

If 11.9 mL of base are required to neutralize 12.8 mL of the acid, what is the molarity of the barium hydroxide solution?

4.

A prevailing pressure of one standard atmosphere will support a column of mercury 760 mm in height. Liquid gallium has a density of 6.20 g/mL. The height of a column of liquid gallium that one standard atmosphere can support is  mm.

The density of liquid mercury is 13.6 g/mL.

5. Liquid methylene bromide has a density of 2.50 g/mL. A barometer is constructed using methylene bromide instead of mercury. If the atmospheric pressure is 0.953atm, what is the height of the methylene bromide column in the barometer in cm?

The density of liquid mercury is 13.6 g/mL.

1)

Balanced chemical equation is:

NaOH + HBr ---> NaBr + H2O

lets calculate the mol of NaOH

volume , V = 13.5 mL

= 1.35*10^-2 L

use:

number of mol,

n = Molarity * Volume

= 0.583*1.35*10^-2

= 7.87*10^-3 mol

According to balanced equation

mol of HBr reacted = (1/1)* moles of NaOH

= (1/1)*7.87*10^-3

= 7.87*10^-3 mol

This is number of moles of HBr

Molar mass of HBr,

MM = 1*MM(H) + 1*MM(Br)

= 1*1.008 + 1*79.9

= 80.908 g/mol

use:

mass of HBr,

m = number of mol * molar mass

= 7.87*10^-3 mol * 80.91 g/mol

= 0.6368 g

mass % of HBr = mass of HBr * 100 / mass of sample

= 0.6368 * 100 / 14.9

= 4.27 %

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