WILL GIVE BRAINIEST IF U ANSWER GOOD If a molecule in the reaction does NOT need to be multiplied, use the number 1
for its coefficient.

tell me what to put in box (1) (2) (3) and (4)

This chemical equation is a little tricky to balance if you don't want to use oxidation numbers, which is assume is the case here.

Explanation:

The first thing to look for in such cases is to balance the elements that are present in odd number on one side of the equation and in even number on the other side of the equation.

In this case, you have

3

atoms of hydrogen on the reactants' side and

2

on the products' side. To balance these out, multiply the ammonia molecule by

2

and the water molecule by

3

2

NH

3

+

O

2

→

NO

+

3

H

2

O

Now, the nitrogen and the oxygen atoms are now unbalanced. Focus on the nitrogen atoms first, and leave the oxygen atoms for last, you'll see why in a minute.

Since you now have

2

atoms of nitrogen on the reactants' side, and only

1

on the products' side, multiply the nitric oxide molecule by

2

to get

2

NH

3

+

O

2

→

2

NO

+

3

H

2

O

Now everything is balanced except the oxygen atoms, since you have

5

on the products' side and only

2

on the reactants' side.

On the reactants' side, oxygen is present as a molecule,

O

2

, which means that you have some elbow room in terms of what stoichiometric coefficient you can assign here.

So, what number multiplied by

2

will give you

5

?

x

⋅

2

=

5

⇒

x

=

5

2

This means that if you multiply the oxygen molecule by

5

2

, you will have

5

oxygen atoms on both sides of the equation.

2

NH

3

+

5

2

O

2

→

2

NO

+

3

H

2

O

Finally, to remove the fractional coefficient, multiply all the species by

2

4

NH

3

+

5

O

2

→

4

NO

+

6

H

2

O

And there you have it, the balanced chemical equation for this reaction.

assuming that your periodic table has eighteen columns, you should highlight   the four elements in the third column from the left of the table.

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the have energy higher than those in the orbital while lower than those of the orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the orbital is therefore filled right after orbitals and before orbitals.  

potassium has electron configuration and calcium . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the and should be highlighted as well.