A chemistry student heated 2.255 g of oxygen gas in the presence of 5.145

Chemistry

2 answers:

m_a_m_a [10]3 years ago

5 0

The answer to your question is 6.21 L

Sergio [31]3 years ago

3 0

Answer:

Explanation:

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1 example of o neutral

katrin [286]

Answer:

NaCl and CO are two neutral compounds as well as these are popular and these Neutral substances are those which have a pH of 7 even they are neither acidic nor basic. The substances which are neither acidic, nor basic and these are called neutral substance. For example distilled water, sugar solution etc. 3.9. 17 votes.

Explanation:

5 0

4 years ago

2Na + 2H2O → 2NaOH + H2 During a laboratory experiment, a certain quantity of sodium metal reacted with water to produce sodium

anyanavicka [17]

Answer : The initial quantity of sodium metal used is 17.25 grams.

Solution : Given,

Volume of $H_2$ gas = 8.40 L

Molar mass of Na metal = 23 g/mole

The Net balanced chemical reaction is,

$2Na+2H_2O\rightarrow 2NaOH+H_2$

At STP, 22.4 L of volume is occupied by 1 mole of $H_2$ gas

so, 8.40 L of volume is occupied by = $\frac{8.40L\times 1mole}{22.4L}$ = 0.375 moles of $H_2$ gas

Now from the above reaction, we conclude that

1 mole of $H_2$ gas produced by the 2 moles of Na metal

0.375 moles of $H_2$ gas produced = $\frac{2moles\times 0.375mole}{1mole}=0.75moles$ of Na metal

The quantity of Na metal used = Moles of Na metal × Molar mass of Na metal = 0.75 moles × 23 g/mole = 17.25 grams

Therefore, the initial quantity of sodium metal used is 17.25 grams.

4 0

3 years ago

Sodium chlorate decomposes into sodium chloride and oxygen gas as seen in the equation below.

Andru [333]

Answer:

${ \rm{2\: moles \: of \: sodium \: chlorate = 2 \: moles \: of \: sodium \: choride}} \\ { \rm{10 \: moles \: of \: sodium \: chlorate = ( \frac{10}{2} \times 2) \: moles}} \\ \\ = { \boxed{ \rm{ \: 10 \: moles \: of \: sodium \: chloride}}} \\$