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Advocard [28]
3 years ago
6

Write and balance the combination reaction for the heating of solid magnesium in the presence of nitrogen gas. In a particular e

xperiment, a 9.27-g sample of N2 completely reacts. What is the mass of magnesium must have been consumed during the reaction if the reaction yield is 100%?
Chemistry
1 answer:
Makovka662 [10]3 years ago
4 0

Answer:

The mass of magnesium that has been consumed, was 6.69 g

Explanation:

The reaction is this one:

3Mg (s)  +  N₂(g) →  Mg₃N₂

3 moles of solid magnesium react with 1 mol of nitrogen, to make 1 mol of magnesium nitride.

If 9.27 grams of nitrogen react, we see that ratio is 1:1, so we make 9.27 grams of nitride.

Mass / Molar mass = Moles

9.27 g / 100.9 g/m = 0.092 moles

If we have 0.092 moles of nitride, ratio between Mg is 1:3 so, the rule of three will be:

1 mol of Nitride was produced by 3 moles of Mg (s)

0.092 moles of nitride were produced by, (0.092 .3)/1 = 0.275 moles

Mass og Mg = 24.3 g/m

Molar mass . Moles = Mass

0.275 m . 24.3g/m = 6.69 g

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A plot of binding energy per nucleon (Eb/ A) versus the mass number (A) shows that nuclei with a small mass number have a small
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a)   1.12 MeV / nucleon

b)   5.62 MeV / nucleon

c)  8.80 MeV / nucleon

d) 8.56 MeV / nucleon

we can conclude that the binding energy has a maximum value for nuclei with a mass around 60

Explanation:

Binding energy = ( Δm * 931.5 ) MeV

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Given that the theoretical mass = 2.0141 u

Actual mass = 1.0078 u + 1.0087 u = 2.0165 u

Δm  = 2.0165 u - 2.0141 u = 2.4 * 10^-3 u

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<u>b) ⁷Li = 3 protons , 4 neutrons = 7 nucleons </u>

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<u>C) ⁶²Ni = 28 protons , 34 neutrons = 62 nucleons </u>

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Actual mass = ( 28 * 1.0078 ) u + ( 34 * 1.0087 ) u

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Actual mass = ( 48 * 1.0078 ) + ( 62 * 1.0087 )

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