Explanation:
From the knowledge of law of multiple proportions,
mass ratio of S to O in SO:
mass of S : mass of O
= 32 : 16
= 32/16
= 2/1
mass ratio of S to O in SO2:
= mass of S : 2 × mass of O
= 32 : 2 × 16
= 32/32
= 1/1
ratio of mass ratio of S to O in SO to mass ratio of S to O in SO2:
= 2/1 ÷ 1/1
= 2
Thus, the S to O mass ratio in SO is twice the S to O mass ratio in SO2.
I think your answer is D. However I’m not 100% sure.
There are six electrons in the covalent bonds.
Two N atoms would be :N:· + ·:N:
An N₂ molecule would be :N:::N: or :N≡N:
This gives each N atom an octet of eight electrons in its valence shell.
Answer:
63.53% yield
Explanation:
The balanced equation for this reaction is 2NaCl + H2O -> 2NaOH +Cl2
First we must find the limiting reactant
From NaCl we can only produce 6.06 grams of Cl2 in <u>theory</u>
From H20 we can only produce 38.995 grams in theory
so we know NaCl is the limiting
% yield is (Actual/Theoretical) x100 so
(3.85/6.06)x100= 63.53% yield
