Answer
pOH = 4.1
Explanation
<em>Given:</em>
pH = 9.9
<em>Required</em>: The concentration of OH-
Solution
pH + pOH = 14
9.9 + pOH = 14
pOH = 14-9.9
pOH = 4.1
Explanation:
The given cell reaction is as follows.
Hence, reactions taking place at the cathode and anode are as follows.
At anode ; Oxidation-half reaction : 
...... (1)
At cathode; Reduction-half reaction : 
....... (2)
Hence, balance the half reactions by multiplying equation (1) by 2 and equation (2) by 3.
Therefore, net cell reaction is as follows.
Net reaction: 
Thus, we can conclude that the overall cell reaction is as follows.
Answer:
Nitrogen (ii) oxide
Explanation:
To know the IUPAC name for NO, we shall determine the oxidation number of N in NO.
NOTE: The oxidation number of oxygen (O) is always – 2.
Thus the oxidation number of N in NO can be obtained as follow:
N + O = 0 (ground state)
N + (– 2) = 0
N – 2 = 0
Collect like terms
N = 0 + 2
N = +2
Thus, the oxidation number of Nitrogen (N) in NO is +2.
Therefore, the IUPAC name for NO is Nitrogen (ii) oxide
Answer:
First, find out how many moles of N2I6 you have. Then convert that to grams.
molar mass N2I6 = 789 g
moles N2I6 = 8.2x1022 molecules N2I6 x 1 mole/6.02x1023 molecules = 1.36x10-1 moles = 0.136 moles
grams N2I6 = 0.136 moles x 789 g/mole = 107 g = 110 g (to 2 significant figures)
