Answer:
M (third main energy level)
Explanation:
The third main energy level bears the first appearance of the 'd' sublevel. The principal quantum number(n) depicts the main energy levels in which an orbital is located. It takes values of n=1,2,3,4,5..... and it can be represented by the shells k,l,m,n.......
The subshells in these main orbitals are represented by s,p,d and f. For the K shell, the principal quantum number is m and its sublevel notations are s,p and d. This is where the d-sublevel first appears.
Answer:
Results from method B is more reliable than method A.
Explanation:
The two method that are used for the analysis produced different results. The first method that is method A gives higher value of the iodine content than the method B.
When 
was added to water, method A showed an increased in the iodine content and it increases with the increase in the amount of .
Where as in the method B, there is no change in the results. Therefore the measurements provided by the method A shows an inference of ion.
The measurement of the iodine content is affected by the presence of the ion in water.
Since in method B there is no change in measurement, it is independent of the presence ion in water.
As higher iodine content is given by method A, so ion must be present in original water that must be interfering the measurement. Hence, method B is more reliable.
Alcohol, alcohol is the only liquid since copper and aluminum are a solid ammonia is a bacteria
Wood, Water, and Neon Gas, because the wood is more of a solid its particles are tightly arranged and water next because it takes the shape of the container it is in and then the neon gas because with a gas the particles are not at all arranged
Answer:
6H2 + P4→ 4PH3
Explanation:
Phosphorus has 4 in it and hydrogen has 3 in it. in order to balance it, we have to put 4 in front of phosphine so that the phosphorus on the product side has an equal amount as to the one on the reactant side.
the only one left to balance is hydrogen and so in order to balance it we put a 6 on h2 because the hydrogen in the product size becomes 12 (4 * 3).
therefore the hydrogen on the reactant side becomes 12 as well (6 * 2)
