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ASHA 777 [7]
3 years ago
6

Balance the following equation: N2 + H2 --> NH3

Chemistry
1 answer:
Marina86 [1]3 years ago
5 0

Answer:

3, 2

Explanation:

Balanced equation:

N2 + 3 H2 = 2 NH3

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Drag each tile to the correct location on the image.
nikdorinn [45]

Answer:

[He]: 2s² 2p⁵.

[Ne]: 3s².

[Ar]: 4s² 3d¹⁰ 4p².

[Kr]: 5s² 4d¹⁰ 5p⁵.

[Xe]: 6s² 4f¹⁴ 5d¹⁰ 6p².

Explanation:

  • Noble elements are used as blocks in writing the electronic configuration of other elements as they are stable elements.

  • [He]:

He contains 2 electrons fill 1s (1s²).

So, [He] can be written before the electronic configuration of 2s² 2p⁵.

  • [Ne]:

Ne contains 10 electrons fill (1s² 2s² 2p⁶).

So, [Ne] can be written before the electronic configuration of 3s².

  • [Ar]:

Ar contains 18 electrons is configured as ([Ne] 3s² 3p⁶).

So, [Ar] can be written before the electronic configuration of 4s² 3d¹⁰ 4p².

  • [Kr]:

Kr contains 36 electrons is configured as ([Ar] 4s² 3d¹⁰ 4p⁶).

So, [Kr] can be written before the electronic configuration of 5s² 4d¹⁰ 5p⁵.

  • [Xe]:

Xe contains 54 electrons is configured as ([Kr] 5s² 4d¹⁰ 5p⁶).

So, [Xe] can be written before the electronic configuration of 6s² 4f¹⁴ 5d¹⁰ 6p².

3 0
3 years ago
Read 2 more answers
If 12.6 grams of iron (III) oxide reacts with 9.65 grams of carbon monoxide to produce 7.23 g of pure iron, what are the theoret
Mariana [72]
The balanced equation is Fe₂O₃ + 3 CO = 2 Fe + 3 CO₂.
Next step is to convert everything to moles.
12.6g Fe₂O₃ x (1 mol Fe₂O₃ / 159.7g Fe₂O₃) = 0.07890 mol Fe₂O₃ 
9.65g CO x (1 mol CO / 28.01g CO) = 0.3445 mol CO
The third step is to determine the limiting and excess reactants.
0.07890 mol Fe₂O₃ x (3 mol CO/1 mol Fe₂O₃) = 0.2367 mol CO
Therefore Fe₂O₃ is the limiting reagent while CO is in excess.

0.07890 mol Fe x (2 mol Fe(s) / 1 mol Fe₂O₃) = 0.1578 mol Fe(s) 
0.1578 mol Fe x (55.84g Fe / mole Fe) = 8.812g Fe is the theoretical yield
%yield = (7.23g / 8.812g) x 100% = 82.0% is the percent yield
3 0
3 years ago
a compound has a molar mass of 129 g/mol if its empirical formula is C2H5N then what is the molecular formula
Elena-2011 [213]

Given :

A compound has a molar mass of 129 g/mol .

Empirical formula of compound is C₂H₅N .

To Find :

The molecular formula of the compound.

Solution :

Empirical mass of compound :

M_e = ( 2 \times 12 ) + ( 5 \times 1 ) + (  1  \times 14 )\\\\M_e = 43\ gram/mol

Now, n-factor is :

n = \dfrac{M}{M_e}\\\\n = \dfrac{129}{43}\\\\n = 3

Multiplying each atom in the formula by 3 , we get :

Molecular Formula, C₆H₁₅N₃

3 0
2 years ago
About how many elements are found on the periodic table? Question 8 options:
krek1111 [17]

the answer is actually 118 look it up

3 0
3 years ago
Why is an alkali metal element like sodium (Na) so reactive?
guapka [62]
Alkali metals are known for being some of the most reactive metals. This is due in part to their larger atomic radii and low ionization energies. They tend to donate their electrons in reactions and often have an oxidation state of +1. These metals are characterized as being extre
3 0
3 years ago
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