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:
option D is correct
D. This solution is a good buffer.
Explanation:
TRIS (HOCH
)
CNH
if TRIS is react with HCL it will form salt
(HOCH)CNH + HCL ⇆ (HOCH)NHCL
Let the reference volume is 100
Mole of TRIS is = 100 × 0.2 = 20
Mole of HCL is = 100 × 0.1 = 10
In the reaction all of the HCL will Consumed,10 moles of the salt will form
and 10 mole of TRIS will left
hence , Final product will be salt +TRIS(9 base)
H = Pk
+ log (base/ acid)
8.3 + log(10/10)
8.3
I believe that the answer is 12 because there is already 3 O molecules and since its in parentheses with 3 outside it that means that there are 3 of those CO molecules meaning that for every 1 CO there will be 3 O’s so 3, four times Is 12
Explanation:
Number of moles(n)=Number of atoms(N)/Avogadro's constant.
Avogadro's constant=6.02×10²³
so we have
n=9.05×10²³/6.02×10²³
n=1.0503moles.
n=mass/molar mass
1.0503=mass/28
mass=1.0503×28
mass=29.4084g
