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AfilCa [17]
3 years ago
9

Hlo guys how are you all ​

Chemistry
2 answers:
kifflom [539]3 years ago
5 0

Answer:

I'm greattt wbu??? :)))

tangare [24]3 years ago
4 0
Could be better, could be worse
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639.232 grams

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A student wants to reclaim the iron from an 18.0-gram sample of iron(III) oxide, which
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m_{Fe}=12.6gFe

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Hello,

In this case, since we have grams of iron (III) oxide whose molar mass is 159.69 g/mol are able to compute the produced grams of iron by using its atomic mass that is 55.845 g/mol and their 2:4 molar ratio in the chemical reaction:

m_{Fe}=18.0gFe_2O_3*\frac{1molFe_2O_3}{159.69gFe_2O_3}*\frac{4molFe_2O_3}{2molFe_2O_3} *\frac{55.845gFe}{1molFe_2O_3} \\\\m_{Fe}=12.6gFe

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What mass of nitrogen is needed to fill an 855 L tank at STP?
scoundrel [369]

Answer:

It takes 1,068.76 grams of nitrogen to fill an 855 L tank at STP.

Explanation:

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C or 273.15 °K are used and are reference values for gases.

On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

So, in this case:

  • P= 1 atm
  • V= 855 L
  • n= ?
  • R= 0.082 \frac{atm*L}{mol*K}
  • T= 273.15 K

Replacing:

1 atm* 855 L= n* 0.082 \frac{atm*L}{mol*K} * 273.15 K

Solving:

n=\frac{1 atm* 855 L}{0.082\frac{atm*L}{mol*K} *273.15 K }

n= 38.17 moles

Being the molar mass of nitrogen N2 equal to 28 g / mol, you can apply the following rule of three: if there are 28 grams in 1 mole, how much mass is there in 38.17 moles?

mass=\frac{38.17 moles*28 grams}{1 mole}

mass= 1,068.76 grams

<u><em> It takes 1,068.76 grams of nitrogen to fill an 855 L tank at STP.</em></u>

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3 years ago
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