A teaspoon of caffeine is <em>NOT</em> deadly as teaspoon of ricin
Let's eliminate these one by one.
The first pair would not be the same, as X would most likely be in group IA, and Y would be in group VIIA, because of their tendency to gain and lose electrons.
The second pair would also violate the same rule, but X would most likely be in group IIA, and Y would most likely be in group VIA.
The third pair would not be the same, as X is most likely in group VIIA, and since Y has eight valence electrons, it is most likely a noble gas.
The final pair has X with atomic number 15, making it phosphorous. Phosphorous wants to gain 3 electrons to have a full octet of 8 outer "valence" electrons, and Y would also like to gain 3 electrons. This means it is possible that the final pair would be in the same group.
Answer:
3.59x10^21 molecules
Explanation:
1mole of a substance contains 6.02x10^23 molecules.
Therefore, 1mole of C8H18 will also contain 6.02x10^23 molecules
1mole of C8H18 = (12x8) +(18x1) = 96 + 18 = 114g.
1mole (i.e 114g) oh C8H18 contains 6.02x10^23 molecules.
Therefore, 0.68g of C8H18 will contain = (0.68 x 6.02x10^23)/114 = 3.59x10^21 molecules
Answer:
NaCl.
Explanation:
In the solution, ZnSe ionizes to 
and 
. Following reaction represents the ionization of ZnSe in solution -
⇄ 
As we want to increase the solubility of ZnSe, we must decrease the concentration of dissociated ions so that the reaction continues to forward direction.
If we add NaCl to this solution, then we have 
and 
in the solution which will be formed by the ionization of NaCl.
Now, in the solution will react with two ions to form 
as follows -
⇄
Due to this reaction the concentration of will decrease in the solution and more ZnSe can be soluble in the solution.
-3 is the answer your looking for
