<u>Answer:</u> The chemical formula is 
<u>Explanation:</u>
Ionization energy is defined as the amount of energy that is required to remove an electron from a chemical species.
The ionization energy equation for the given values follow:
From the values of ionization energy above, it can be seen that the ionization energy increases as every succeeding electron is removed.
Second ionization energy is a little higher than the first one but there is a huge amount of difference between the third and second ionization energy.
This implies that the ion formed during second ionization energy has a stable configuration and it requires a humongous amount of heat to release the third electron.
Hence, the ion formed will be 
Sulfite ion is a polyatomic ion having a chemical formula of 
An ionic compound is formed between the two ions and the chemical compound formed between the two will have a formula of
Hence, the chemical formula is
<span>The correct answer is 3: 0.10 M K3PO4(aq). The higher the concentration (or molarity), the higher the boiling point. Thus, the solution with the most moles will have the highest boiling point. SO4, PO4, and NO3 are all polyatomic ions, so by definition, they only have one mole. In K3PO4, K3 has 3 moles and PO4 has 1 mole, meaning all together it has 4 moles (more than any of the other options).</span>
pH=6.98
Explanation:
This is a very interesting question because it tests your understanding of what it means to have a dynamic equilibrium going on in solution.
As you know, pure water undergoes self-ionization to form hydronium ions, H3O+, and hydroxide anions, OH−.
2H2O(l]⇌H3O+(aq]+OH−(aq]→ very important!
At room temperature, the value of water's ionization constant, KW, is equal to 10−14. This means that you have
KW=[H3O+]⋅[OH−]=10−14
Since the concentrations of hydronium and hydroxide ions are equal for pure water, you will have
[H3O+]=√10−14=10−7M
The pH of pure water will thus be
pH=−log([H3O+])
pH=−log(10−7)=7
Now, let's assume that you're working with a 1.0-L solution of pure water and you add some 10
Answer: -
0.1 ml of bleach should be added to each liter of test solution.
Explanation:-
Let the volume of bleach to be added is B ml.
Density of stock solution = 1.0 g/ml
Mass of stock solution = Volume of stock x density of stock
= B ml x 1.0 g/ml
= B g
Amount of NaOCl in this stock solution = 5% of B g
= 
x B g
= 0.05 B g
Now each test solution must be added 5 mg/l NaOCl.
Thus each liter of test solution must have 5 mg.
Thus 0.05 B g = 5 mg
= 0.005 g
B = 
= 0.1
Thus 0.1 ml of bleach should be added to each liter of test solution.
