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Dmitriy789 [7]
3 years ago
10

Ok so N2+O2=2NO and my delta h is 180.5 kJ I need to know. If I react 10 g of nitrous gas with excess oxygen how many grams of N

itrogen monoxide will be produced
Chemistry
1 answer:
Westkost [7]3 years ago
4 0

Answer:

21.6 grams

Explanation:

  • N₂ + O₂ → 2NO

First we <u>convert 10 g of N₂ into moles</u>, using its <em>molar mass</em>:

  • 10 g ÷ 28 g/mol = 0.36 mol N₂

Then we <u>convert 0.36 moles of N₂ into moles of NO</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:

  • 0.36 mol N₂ * \frac{2molNO}{1molN_2} = 0.72 mol NO

Finally we <u>convert 0.72 NO moles into grams</u>, using its <em>molar mass</em>:

  • 0.72 mol NO * 30 g/mol = 21.6 g
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Answer:

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Explanation:

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3 years ago
(a) (1)
Elis [28]

Explanation:

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M(g)  ®  M+(g)  +  e-

It is possible to remove more electrons from most elements, so this quantity is more precisely known as the first ionization energy, the energy to go from neutral atoms to cations with a 1+ charge.  The second ionization energy is the energy that is required to remove a second electron, to form 2+ cations from 1+ cations:

M+(g)  ®  M2+(g)  +  e-

The third ionization energy is the energy required to form 3+ cations:

M2+(g)  ®  M3+(g)  +  e-

and so on.  Ionization energies are always positive numbers, because energy must be supplied (an endothermic energy change) to separate electrons from atoms.  The second ionization energy is always larger than the first ionization energy, because it requires even more energy to remove an electron from a cation than it is from a neutral atom.

The first ionization energy varies in a predictable way across the periodic table.  The ionization energy decreases from top to bottom in groups, and increases from left to right across a period.  Thus, helium has the largest first ionization energy, while francium has one of the lowest.

From top to bottom in a group, orbitals corresponding to higher values of the principal quantum number (n) are being added, which are on average further away from the nucleus.  Since the outermost electrons are further away, they are less strongly attracted by the nucleus, and are easier to remove, corresponding to a lower value for the first ionization energy.From left to right across a period, more protons are being added to the nucleus, but the number of electrons in the inner, lower-energy shells remains the same.  The valence electrons feel a higher effective nuclear charge — the sum of the charges on the protons in the nucleus and the charges on the inner, core electrons.  The valence electrons are therefore held more tightly, the atom decreases in size (see atomic radius), and it becomes increasingly difficult to remove them, corresponding to a higher value for the first ionization energy.

 

The following charts illustrate the general trends in the first ionization energy:

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8 0
3 years ago
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Answer:

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Explanation:

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                                           \|/

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3 years ago
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Answer:

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