and 
.
Assuming complete decomposition of both samples,
First compound:
; 
of the first compound would contain
Oxygen and mercury atoms seemingly exist in the first compound at a 
ratio; thus the empirical formula for this compound would be where the subscript "1" is omitted.
Similarly, for the second compound
; 
of the first compound would contain
and therefore the empirical formula
.
Answer:
A) SiO2 is the limiting reactant
B) Theoretical yield= 72333.3g
C) % yield =91.5%
Explanation:
SiO2(s) + 2C(s) --------------> Si(s) + 2CO(g)
n(SiO2)= 155000/60 = 2583.33 mols
n(C)= 79000/12= 3291.66 mols
a)SiO2 is the limiting reactant
According to the balanced reaction equation,
60g of SiO2 produced 28g of SiO2
155000g of SiO2 will produce 155000×28/60= 72333.3g
Therefore theoretical yield of Si= 72333.3g
% yield= 66200/72333.3×100/1 =91.5%
The answer is both B and C
positive chromic acid test is indicated by disappearance of orange colour from chromic ions and appearance of blue-green color from Chromium (iii) ion (reduction of chromium ion from CrO4 - to Cr3+)
positive chromic test indicated functional groups that can be oxidized.
cyclohexanol can be oxidized to become cyclohexanone
and pentan-3-ol can be oxidized to become pentan-3-one
hence both B and C will show positive chromic acid test
A) tert butanol although contains alcohol functional group, cannot be further oxidized as it is a tertiary alcohol
Answer:
Name Atomic Number Electron Configuration Period 1 Hydrogen 1 1s1 Helium 2 1s2 Period 2 Lithium 3 1s2 2s1 Beryllium 4 1s2 2s2 Boron 5 1s2 2s22p1 Carbon 6 1s2 2s22p2 Nitrogen 7 1s2 2s22p3 Oxygen 8 1s2 2s22p4 Fluorine 9 1s2 2s22p5 Neon 10 1s2 2s22p6 Period 3 Sodium 11 1s2 2s22p63s1 Magnesium 12 1s2 2s22p63s2 Aluminum 13
