PbO2
You have to take the mass of lead in the problem, and divide by the molar mass.
When you do the same with oxygen, you get a number about twice as large as when you divide the mass of lead by the molar mass of lead. This means that the simplest formula would be PbO2
<u>Answer:</u> The equilibrium constant for 
equation is 
<u>Explanation:</u>
The given chemical equation follows:
The value of equilibrium constant for the above equation is 
Calculating the equilibrium constant for the given equation:
The value of equilibrium constant for the above equation will be:
Hence, the equilibrium constant for equation is
The question is incomplete, the complete question is;
Which of the following is most likely a heavier stable nucleus? (select all that apply) Select all that apply: A nucleus with a neutron:proton ratio of 1.05 A nucleus with a A nucleus with a neutron:proton ratio of 1.49 The nucleus of Sb-123 A nucleus with a mass of 187 and an atomic number of 75
Answer:
A nucleus with a A nucleus with a neutron:proton ratio of 1.49
A nucleus with a mass of 187 and an atomic number of 75
Explanation:
The stability of a nucleus depends on the number of neutrons and protons present in the nucleus. For many low atomic number elements, the number of protons and number of neutrons are equal. This implies that the neutron/proton ratio = 1
Elements with higher atomic number tend to be more stable if they have a slight excess of neutrons as this reduces the repulsion between protons.
Generally, the belt of stability for chemical elements lie between and N/P ratio of 1 to an N/P ratio of 1.5.
Two options selected have an N/P ratio of 1.49 hence they are heavy stable elements.
Answer : The normality of the solution is, 30.006 N
Explanation :
Normality : It is defined as the number of gram equivalent of solute present in one liter of the solution.
Mathematical expression of normality is:
or,
First we have to calculate the equivalent weight of solute.
Molar mass of solute 
= 94.97 g/mole
Now we have to calculate the normality of solution.
Therefore, the normality of the solution is, 30.006 N
The oxidation numbers of nitrogen in NH3, HNO3, and NO2 are, respectively: -3, -5, +4 +3, +5, +4 -3, +5, -4 -3, +5, +4
Evgesh-ka [11]
In NH3 , let oxidation number of N be x
x + (+1)3 = 0
x = -3
In HNO3 , let oxidation number of N be x
1 + x + (-2)3 = 0
x = +5
In NO2 , let oxidation number of N be x
x + (-2)2 = 0
x = +4
