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Vaselesa [24]
3 years ago
9

Which of these compounds would most likely be found in a deposit of natural gas?

Chemistry
2 answers:
Valentin [98]3 years ago
8 0
The correct answer for the given question above would be option A. The compound that would most likely be found in a deposit of natural gas is CH4 or METHANE. Methane is the main constituent of natural gas. It is<span> a colorless, odorless gas with a wide distribution in nature. Hope this answers your question.</span>
ANTONII [103]3 years ago
4 0

Answer: It A. CH4

Explanation:

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True/False: All of the elements combine in a variety of ways to make up all living and all nonliving things
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Susan has a beaker with one mole of water and another beaker filled with one mole of hydrogen chloride. Compare the number of mo
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Ratio is 1:1 ; they have same number of molecules.

Explanation:

One mole of any compound have same number of molecules, irrespective of their chemistry and composition, i.e. 6.022x10^23 . This number is known as Avogadro's Number.

So the beaker with one mole of water contains = 6.022x10^23 molecules ;

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Fe2O3(s) + 3CO(g) ---&gt; 2Fe(l) + 3CO2(g) Steve inserts 450. g of iron(III) oxide and 260. g of carbon monoxide into the blast
baherus [9]

Answer: Theoretical yield is 313.6 g and the percent yield is, 91.8%

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Fe_2O_3=\frac{450}{160}=2.8moles

\text{Moles of} CO=\frac{260}{28}=9.3moles

Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(l)+3CO_2(g)

According to stoichiometry :

1 mole of Fe_2O_3 require 3 moles of CO

Thus 2.8 moles of Fe_2O_3 will require=\frac{3}{1}\times 2.8=8.4moles  of CO

Thus Fe_2O_3 is the limiting reagent as it limits the formation of product and CO is the excess reagent.

As 1 mole of Fe_2O_3 give = 2 moles of Fe

Thus 2.8 moles of Fe_2O_3 give =\frac{2}{1}\times 2.8=5.6moles of Fe

Mass of Fe=moles\times {\text {Molar mass}}=2.6moles\times 56g/mol=313.6g

Theoretical yield of liquid iron = 313.6 g

Experimental yield = 288 g

Now we have to calculate the percent yield

\%\text{ yield}=\frac{\text{Actual yield }}{\text{Theoretical yield}}\times 100=\frac{288g}{313.6g}\times 100=91.8\%

Therefore, the percent yield is, 91.8%

6 0
4 years ago
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