I would use a conversation calculator
Based on the assumption that the reaction involves N and O to produce NO, if 25.0 g of NO are produced, the amount of N gas used would be 11.66 grams
<h3>Stoichiometric calculation</h3>
From the equation of the reaction:
N + O ---------> NO
Mole ratio of N to NO is 1:1
Mole of 25.0 g of NO = 25/30.01 = 0.833 moles
Equivalent mole of N = 0.833 moles
Mass of 0.833 moles N = 0.833 x 14 = 11.66 grams
More on stoichiometric calculations can be found here: brainly.com/question/8062886
Answer:
Sc
:
4s
2
3d
1
Y
:
l
5s
2
4d
1
La
:
6s
2
5d
1
Ce
:
6s
2
4f
1
5d
1
Gd
:
6s
2
4f
7
5d
1
Lu
:
6s
2
4f
14
5d
1
Ac
:
7s
2
6d
1
Pa
:
7s
2
5f
2
6d
1
U
:
l
7s
2
5f
3
6d
1
Np
:
7s
2
5f
4
6d
1
Cm
:
7s
2
5f
7
6d
1
b. s2p3
he pnictogens:
N
l
:
2s
2
2p
3
P
l
:
3s
2
3p
3
As
:
4s
2
3d
10
4p
3
Sb
:
5s
2
4d
10
5p
3
Bi
:
6s
2
4f
14
5d
10
6p
3
Mc
:
7s
2
5f
14
6d
10
7p
3
c.The noble gases:
Ne
:
2s
2
2p
6
Ar
:
3s
2
3p
6
Kr
:
4s
2
3d
10
4p
6
Xe
:
5s
2
4d
10
5p
6
Rn
:
6s
2
4f
14
5d
10
6p
6
Og
:
7s
2
5f
14
6d
10
7p
6
Explanation: From the periodic tables we can drive elements with the electronic configuration
Sc
:
4s
2
3d
1
Y
:
l
5s
2
4d
1
La
:
6s
2
5d
1
Ce
:
6s
2
4f
1
5d
1
Gd
:
6s
2
4f
7
5d
1
Lu
:
6s
2
4f
14
5d
1
Ac
:
7s
2
6d
1
Pa
:
7s
2
5f
2
6d
1
U
:
l
7s
2
5f
3
6d
1
Np
:
7s
2
5f
4
6d
1
Cm
:
7s
2
5f
7
6d
1
b. s2p3
he pnictogens:
N
l
:
2s
2
2p
3
P
l
:
3s
2
3p
3
As
:
4s
2
3d
10
4p
3
Sb
:
5s
2
4d
10
5p
3
Bi
:
6s
2
4f
14
5d
10
6p
3
Mc
:
7s
2
5f
14
6d
10
7p
3
c.The noble gases:
Ne
:
2s
2
2p
6
Ar
:
3s
2
3p
6
Kr
:
4s
2
3d
10
4p
6
Xe
:
5s
2
4d
10
5p
6
Rn
:
6s
2
4f
14
5d
10
6p
6
Og
:
7s
2
5f
14
6d
10
7p
6
Answer:
12
Explanation:
If GT bisects BN at point G, then point G would be in the middle of BN. So, since Bg and NG have to be equal and add up to 24, then the answer is 12.
3Ba + N2 -----> Ba3N2
Ba is in the second group and metal, its charge = +2 in the compound.
N is non-metal, it is in 15th group, so it has 5 electrons on the last level, so it can take 3 electrons more(8-5=3). N has charge (-3) in compound.
Ba and N form compound Ba3N2.