Na₂O is an ionic compound. Na is in group 1 which means it has one electron in the outer shell. O is in group 6 so it has 6 electrons in the outer shell.
To become stable atoms need to either lose the electrons in the outer shell or gain electrons to gain a complete outer shell configuration.
To become stable Na loses its outer electron and becomes positively charged. It only loses one electron, so net charge is +1. O to become stable gains 2 electrons to complete its outer shell and becomes negatively charged. since it gains 2 electrons, the net charge is -2.
cation - Na⁺
anion - O²⁻
O gains 2 electrons but Na can give only one electron, therefore 2 Na⁺ ions are required.
the compound can be written by exchanging the charges
ions Na⁺ O²⁻
charge +1 -2
exchange Na₂O
Correct answer is <span>Na has +1 charge and O has -2 charge</span>
Solution :
Time (sec) Volume of NaOH (mL)
339 26.23
1242 27.80
2745 29.70
4546 3.81
39.81
Now the example of the first order kinetics w.r.t volumetric analysis is :

Here, 

= volume at time 0 = 0
Since the interval is not constant, we take the time interval as


= 1402.3333
≈ 1402 seconds


= 0.001643 x 0.52045
= 0.00082

Therefore, the first order rate constant is k
.
Carbon dioxide is a carbon atom and two oxygen atoms, therefore CO2
For the others, they are hydrocarbons.
The first part of the name is determined by how many carbon atoms there are. The second part, is by the type. Alcohol, Alkane, Alkene, alkynes, acid, esters, amides.
Answer: <span>A-Ce is oxidized because it is losing electrons and Cu is reduced because it is gaining electrons</span><span>.
</span>There are two reactions in the equation, oxidation and reduction. A molecule that oxidized will lose electrons while the molecule that reduced will gain electrons. In this case, Cu2+ changed into Cu which means its oxidation number reduced from +2 into 0. Ce oxidation number increased from 0 into +3