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3 years ago

_H2 + _02 → _H20 Need help

Chemistry

1 answer:

Dominik [7]3 years ago

8 0

Answer:

h + O2

cross over valencies
h20

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4 years ago

Which element would chemically react with oxygen

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3 years ago

A student obtains a 10.0g sample of a white powder labeled as BaCl2. After completely dissolving the powder in 50.0mL of distill

LekaFEV [45]

Answer:

Whether barium chloride solution was pure

Explanation:

We may answer whether barium chloride was pure. The sequence of this experiment might be depicted by the following balanced chemical equations:

$BaCl_2 (aq)\rightarrow Ba^{2+} (aq) + 2 Cl^- (aq)$

$Ba^{2+} (aq) + SO_4^{2-} (aq)\rightarrow BaSO_4 (s)$

Having a total sample of 10.0 grams, we would firstly find the mass percentage of barium in barium chloride:

$\omega_{Ba^{2+}} = \frac{M_{Ba}}{M_{BaCl_2}} = \frac{137.327 g/mol}{208.23 g/mol\cdot 100\% = 65.95 \%$

This means in 10.0 g, we have a total of:

$m_{Ba^{2+}} = 10.0 g\cdot 0.6595 = 6.595 g$ of barium cations.

The precipitate is then formed and we measure its mass. Having its mass determined, we'll firstly find the percentage of barium in barium sulfate using the same approach:

$\omega_{Ba} = \frac{137.327 g/mol}{233.38 g/mol}\cdot 100\% = 58.84 \%$

Multiplying the mass we obtained by the fraction of barium will yield mass of barium in barium sulfate. Then:

if this number is equal to 6.595 g, we have a pure sample of barium chloride;
if this number is lower than 6.595 g, this means we have an impure sample of barium chloride, as we were only able to precipitate a fraction of 6.595 g.

3 0

3 years ago