The grams of N2 that are required to produce 100.0 l of NH3 at STP
At stp 1moles = 22.4 l. what about 100.0 L of NH3
= 100 / 22.4 lx1 moles = 4.46 moles of NH3
write the reacting equation
N2+3H2 =2NH3
by use of mole ratio between N2 to NH3 which is 1:2 the moles of N2 =4.46/2 =2.23 moles of N2
mass = moles x molar mass
= 2.23moles x 28 g/mol = 62.4 grams
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Answer:
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Explanation:
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If it is 60 Celsius that would conver to fare height by means of this equation; (1.8*60)+32°F
Which would come out to.... 140° Fahrenheit... Hardly seems like chilly conditions.
The amount in grams of Al₂O₃ produced is approximately 6.80 g.
Aluminium reacts completely with oxygen(air) to produce Al₂O₃. The reaction can be represented with a chemical equation as follows:
AL + O₂ → Al₂O₃
Let's balance it
4AL + 3O₂ → 2Al₂O₃
4 moles of Aluminium reacts with 3 moles of Oxygen molecules to produce 2 moles of Aluminium oxide. Therefore,
Since, aluminium reacts completely, it is the limiting reagent in the reaction. Therefore,
Atomic mass of AL = 27 g
Molar mass of Al₂O₃ = 101.96 g/mol
4(27 g) of AL gives 2(101.96 g) of Al₂O₃
3.6 g of AL will give ?
cross multiply
mass of Al₂O₃ produced = 3.6 × 203.92 / 108 = 734.112 / 108 = 6.797
mass of Al₂O₃ produced = 6.80 g.
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