Answer : The enthalpy change during the reaction is -6.48 kJ/mole
Explanation :
First we have to calculate the heat gained by the reaction.

where,
q = heat gained = ?
m = mass of water = 100 g
c = specific heat = 
= final temperature = 
= initial temperature = 
Now put all the given values in the above formula, we get:


Now we have to calculate the enthalpy change during the reaction.

where,
= enthalpy change = ?
q = heat gained = 23.4 kJ
n = number of moles barium chloride = 

Therefore, the enthalpy change during the reaction is -6.48 kJ/mole
Answer:
The possible structures are ketone and aldehyde.
Explanation:
Number of double bonds of the given compound is calculated using the below formula.

=Number of double bonds
= Number of carbon atoms
= Number of hydrogen atoms
= Number of nitrogen atoms
The number of double bonds in the given formula - 

The number of double bonds in the compound is one.
Therefore, probable structures is as follows.
(In attachment)
The structures I and III are ruled out from the probable structures because the signal in 13C-NMR appears at greater than 160 ppm.
alkene compounds I and II shows signal less than 140 ppm.
Hence, the probable structures III and IV are given as follows.
The carbonyl of structure I appear at 202 and ketone group of IV appears at 208 in 13C, which are greater than 160.
Hence, the molecular formula of the compound
having possible structure in which the signal appears at greater than 160 ppm are shown aw follows.
Add the Pressure of neon and argon that is 0.68 +0.35= 1.03
Total pressure that is 1.25 -1.03=0.22 atm
The ionic equation is as below
Ca^2+(aq) + SO4^2-(aq) ---> CaSO4(s)
EXPLANATION
K2SO4(aq) +Cai2(aq) ---> CaSO4(s) + Ki (aq)
ionic equation
= 2K^+(aq) + SO4^2-(aq) + Ca^2+(aq) + 2i^-(aq) --->CaSO4(s) + 2K^+(aq) +2 i^-(aq)
cancel the spectator ions that is 2k^+ and 2i^-
The net ionic equation is therefore
= Ca^2+(aq) + SO4^2-(aq) ----> CaSO4(s)
Answer:- 6984 kJ of heat is produced.
Solution:- From given information, 1367 kJ of heat is produced by the combustion of 1 mole of ethanol. We are asked to calculate the heat produced by the combustion of 235.0 g of ethanol.
Let's convert given grams to moles and multiply by the heat produced by one mole of ethanol to get the total heat produced. Molar mass of ethanol is 46 grams per mole. The set will be:

= 6984 kJ
So, 6984 kJ of heat is produced by the combustion of 235.0 g of liquid ethanol.