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drek231 [11]
3 years ago
6

guys please help me WILL MARK BARINLIEST If you have 5.7 x 10^24 molecules of O2 at STP, how many liters of O2 do you have? must

show all work
Chemistry
1 answer:
zmey [24]3 years ago
5 0

Answer:

212.8 dm^3 or L

Explanation:

1 mole of any sub=6.02×10^23 molecules

X mole of O2=5.7×10^24 molecules

X mole=5.7×10^24/6.02×10^23

=9.5 mole

1 mole of any gas at stp=22.4 dm^3

Therefore, 9.5 mole of O2 will be 22.4×9.5

=212.8 dm^3

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Answer : The correct expression is, (150 × 3 × 31.998) ÷ (232.29 × 1) grams

Explanation :

First we have to calculate the moles of Zn(ClO_3)_2

\text{ Moles of }Zn(ClO_3)_2=\frac{\text{ Mass of }Zn(ClO_3)_2}{\text{ Molar mass of }Zn(ClO_3)_2}=\frac{150g}{232.29g/mole}=\frac{150}{232.29}mole

Now we have to calculate the moles of O_2

The balanced chemical reaction is,

Zn(ClO_3)_2\rightarrow ZnCl_2+3O_2

From the balanced chemical reaction we conclude that,

As, 1 mole of Zn(ClO_3)_2 react to give 3 mole of O_2

So, \frac{150}{232.29} moles of Zn(ClO_3)_2 react to give \frac{150}{232.29}\times \frac{3}{1} moles of O_2

Now we have to calculate the mass of O_2

\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2

\text{ Mass of }O_2=(\frac{150}{232.29}\times \frac{3}{1})mole\times (31.998g/mole)=\frac{150\times 3\times 31.998}{232.29\times 1}g

Therefore, the correct expression for the mass of oxygen gas formed is, (150 × 3 × 31.998) ÷ (232.29 × 1) grams

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3 years ago
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Rasheed calculates that his chemical reaction should produce 4 moles of product, but when he does the experiment, he gets only 3
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Answer:

The answer to your question is 75%

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