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Tpy6a [65]
3 years ago
12

A] 1 N2 + 3 H2 → 2 NH3

Chemistry
1 answer:
Vilka [71]3 years ago
8 0

I haven't done these in some time, so I'm not sure if they are 100% right.

1) 6.3 moles of H2( 2 mol of NH3 / 3  mol of H2)= 4.2 mol of NH3

6.3 moles of H2( 2 mol of NH3/ 3 mol of H2)(17.04 g of NH3/1 mol NH3)= 71.57 g of NH3

2) 2.5 moles of N2(2 mol of NH3/1 mol of N2)= 5 moles of NH3

2.25 moles of N2(2 mol of NH3/ 1 mol of N2)(6.02x10^23 particles/ 1 mol of NH3)= 3.01x10^24 particles of NH3

3) 425 g of NH3(1 mol of NH3/17.04 g NH3)= 24.9 moles of NH3

425 g of NH3(1 mol of NH3/17.04 g of NH3)(1 mol of N2/2 mol of NH3)(28.02 g of N2/1 mol N2)= 349 g of N2

425 g of NH3(1 mol of NH3/17.04 g of NH3)(1 mol of N2/2 mol of NH3)= 12.5 mol of N2

4) 10 moles NH3(3 moles of H2/2 moles of NH3)= 15 moles H2

10 moles NH3(3 mol of H2/2 mol of NH3)(2.02 g of H2/1 mol of H2)= 30.3 g of H2

30.3 g = .0303 liters of H2

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If the energy difference between two electronic states is 214.68 kJ / mol , calculate the frequency of light emitted when an ele
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{\qquad\qquad\huge\underline{{\sf Answer}}}

Here we go ~

Energy difference btween the two electronic states can be expressed as :

{ \qquad \sf  \dashrightarrow \: \Delta E = h\nu}

[ h = planks constant,{\: \nu }= frequency ]

\qquad \sf  \dashrightarrow \:214.68 = 39.79 \times 10 {}^{ - 14}  \times  \nu

\qquad \sf  \dashrightarrow \: \nu =  \cfrac{214.68}{39.79 \times 10 {}^{ - 4} }

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