I believe it would weigh less if it’s blown up with helium because helium is lighter than air
No it is not
hope this helped
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:
113.69°k
Explanation:
V1=85L of helium V2=32L
T1= 29°C +273= 302°K T2=?
T2=<u>TIV2</u>
V1
T2=<u>(302)(32)</u>= <u>9664</u>
85 85
T2= 113.69°K
