Answer:
a) The formal charge on N is 0 in both species.
Explanation:
The formal charge is calculated using the formular;
FC = V-N-B/2
Where;
V = Number of Valence electrons
N = number of nonbonding valence electrons
B = total number of electrons shared in bonds
In NO2-;
The formal charge of N is given as;
FC = 5 - 2 - 6/2
FC = 0
In HNO2
The formal charge of N is given as;
FC = 0
The correct option is;
a) The formal charge on N is 0 in both species.
Answer:
- 5.15×10²⁴ molecules of sulfur dioxide
- 3.63×10²³ molecules of carbon monoxide
- 6.02×10²³ molecules of ammonia
Explanation:
We begin from the relation that 1 mol of molecules contains NA of molecules
NA = 6.02×10²³
Now, we make rules of three:
1 mol has 6.02×10²³ molecules, therefore:
8.55 moles of SO₂ must have (8.55 . NA) / 1 = 5.15×10²⁴ molecules of dioxide
0.603 moles of CO must have (0.603 . NA) / 1 = 3.63×10²³ molecules of monoxide
Avogadro's Number of molecules of NH₃ are 6.02×10²³ molecules of ammonia
The Nassau Din beat the the star has.
It may turn into a black hole if it has a high enough mass.
Answer:
_5_ AsO2−(aq) + 3 Mn2+(aq) + _2_ H2O(l) → _5_ As(s) + _3_ MnO4−(aq) + _4_ H+(aq)
Explanation:
Step 1:
The unbalanced equation:
AsO2−(aq) + 3 Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)
Step 2:
Balancing the equation.
AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)
The above equation can be balanced as follow:
There are 3 atoms of Mn on the left side of the equation and 1 atom on the right side. It can be balance by putting 3 in front of MnO4− as shown below:
AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)
There are 12 atoms of O on the right side and a total of 3 atoms on the left side. It can be balance by putting 5 in front of AsO2− and 2 in front of H2O as shown below:
5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)
There are 4 atoms of H on the left side and 1 atom on the right side. It can be balance by putting 4 in front of H+ as shown below:
5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + 4H+(aq)
There are 5 atoms of As on the left side and 1 atom on the right side. It can be balance by putting 5 in front of As as shown below:
5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → 5As(s) + 3MnO4−(aq) + 4H+(aq)
Now the equation is balanced
