Given :
A 3.82L balloon filled with gas is warmed from 204.9K to 304.8 K.
To Find :
The volume of the gas after it is heated.
Solution :
Since, their is no information about pressure in the question statement let us assume that pressure is constant.
Now, we know by ideal gas equation at constant pressure :

Hence, this is the required solution.
To convert 78.1 g of water at 0° C to Ice at -57.1°C; we can do it in steps;
1. Water at 0°C to ice at 0°C
The heat of fusion of ice is 334 J/g;
Heat = 78.1 × 334 = 26085.4 Joules
2. Ice at 0°C to -57.1°C
Specific heat of ice is 2.108 J/g
Heat = 78.1 × 2.108 J/g = 164.6348 Joules
Thus the total heat energy released will be; 26085.4 + 164.6348
= 26250.0348 J or 26.250 kJ
B and c...will lose electron(s) in forming an Ion.
P is an Anion
b..Fe. and c...Pb form Cations (+) by losing electrons.
d. Se is an Anion.
<u>U</u> <u>VORBELLO</u> <u>FRANÇAIS</u><u>?</u><u>?</u><u>?</u>
Answer:
The reactive nucleophile is Ketone.
Explanation:
In organic chemistry, The process of acid - catalyzed aldol condensation starts from when ketone (or any aldehyde) is converted to an -enol, after which it attacks another ketone/aldehyde that has already been activated by parbonyl oxygen protonation.
The process of this is that first of all the ketone undergoes tautomerization to form -enol. Thereafter, the other carbonyl will undergo protonation which makes the carbon activated towards attack. Now, the nucleophilic enol will be added to the carbonyl in a [1,2]-addition reaction and we will now use deprotonation to obtain the neutral Aldol product.
Now, since only the ketone can produce an -enol, thus it is the nucleophile as aldehydes are better electrophiles