Answer:
c. HF can participate in hydrogen bonding.
Explanation:
<u>The boiling points of substances often reflect the strength of the </u><u>intermolecular forces</u><u> operating among the molecules.</u>
If it takes more energy to separate molecules of HF than of the rest of the hydrogen halides because HF molecules are held together by stronger intermolecular forces, then the boiling point of HF will be higher than that of all the hydrogen halides.
A particularly strong type of intermolecular attraction is called the hydrogen bond, <em>which is a special type of dipole-dipole interaction between the hydrogen atom in a polar bond</em>, such as N-H, O-H, or F-H, and an electronegative O, N, or F atom.
Number of moles in the K2SO4 sample
= (16/1000)*1.04= 0.01664 mol
Number of moles in the Ba(NO3)2 sample
= (14.3/1000*0.880)= 0.01258 mol
Since the reaction is a 1:1 ratio between the two reactants, the limiting reagent is the one containing a smaller number of moles, namely Ba(NO3)2.
The molecular mass of BaSO4 is 137.3+(32.06+4*16.00)=233.4
Therefore the theoretical yield of Barium Sulphate is
233.4*0.01258=2.937 g
Actual yield = 2.60 g (given)
Therefore the percentage yield = 2.60/2.937=88.54%
Answer:
1. the limiting reagent is Barium Nitrate (Ba(NO3)2)
2. the theoretical yield is 2.94 g
3. the percentage yield is 88.5%
I apologize for the mistake previous to this update.
Answer:
i think its a liquid..... im not that sure
Basically since there’s 2 hydrogen’s there will be two H’s on either side of your other element. And Se is in group 6 which means it has 6 valence electrons. When you combine 6 and 2 from the hydrogen you get 8. You then should place 8 dots around Se two on each side.
So something like H Se(with 8 dogs around it) H
put cool water over it not cold and cover the burrn
