Explanation:
The given data is as follows.
Moles of propylene = 100 moles, 
= 300 K
= 800 K, 
, 
of propylene = 100 J/mol
Now, we assume the following assumptions:
Since, it is a compression process therefore, work will be done on the system. And, work done will be equal to the heat energy liberating without any friction.
W = 
= 
= 5 MJ
Thus, we can conclude that a minimum of 5 MJ work is required without any friction.
mass of proton mp=1.672621898e-27 kg and 1 amu=1.66053904e-27kg so it’s just divition (1.672621898e-27 / 1.66053904e-27) =1.007276 amu
Answer:
186.3g
Explanation:
4.5moles of K₂CO₃ is in 1000ml
? moles of K₂CO₃ is in 300 ml
(4.5 × 300)/ 1000 = 1.35 moles of K₂CO₃
1 mole of K₂CO₃ = (39 × 2) + 12 + (16 × 3) = 78 + 12 + 48 = 138g
1.35 moles of K₂CO₃ = ?
= (1.35 × 138)/1 = 186.3g
Given data:
Volume of HCl = 14.22 ml
Molarity of HCl = 2.97 M
mmoles of HCl = 14.22 * 2.97 = 42.2 mmoles
Volume of NaOH = 5.00 ml
Molarity of NaOH = 0.1055 M
mmoles of NaOH = 5.00 *.1055 = 0.5275 mmoles
Since HCl and NaOH combine in a 1:1 ratio
# moles of NaOH = # moles of excess HCl that is neutralized = 0.5275 moles
Now, the total moles of HCl taken = # mmoles HCl neutralized by antacid + # mmoles of excess HCl
42.2 = mmoles HCl neutralized by antacid + 0.5275
Therefore,
mmoles of HCl neutralized by antacid = 42.2 - 0.5275 = 41.6725 mmoles = 41.7 mmoles
By how thick or slim it is
