The correct option would be 3.
Only thermal energy changes
Hope this helps you
Brainliest would be appreciated
-AaronWiseIsBae
Answer: n=15.56moles
Explanation:
PV = nRT
where
P is pressure in atmospheres
V is volume in Liters
n is the number of moles of the gas
R is the ideal gas constant = given as (0.0821L -atm/k-mol
PV = nRT
n= PV/RT
n= (1.5 X 230)/ (0.0821 X 270)
n= 15.56 moles
Answer:
The enthalpy of reaction for the reaction of chlorine with ozone is -162.5 kJ.
Explanation:
..[1]
..[2]
..[3]
The enthalpy of reaction for the reaction of chlorine with ozone can be calculated by using Hess's law:
[2] - [1] = [3]
The enthalpy of reaction for the reaction of chlorine with ozone is -162.5 kJ.
<span>A chemist adds 155.0ml of a 4.10 X 10^-5 mmol/L of a zinc oxalate (ZnC2O4)solution to a reaction flask. Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask.
1mmol = 10^-3 mol
Therefore 4.10*10^-5mmol = 4.10*10^-8mol
molar mass ZnC2O4 = 65.39+(2*12.011)+(4*15.99) = 153.372g/mol
You have 4.10*10^-8 mol/litre =153.372 * 4.10*10^-8 = 6.29*10^-6 grams / litre (* see below)
But you have 155ml. Mass of ZnC2O4 = 155/1000*6.29*10^-6 g
Mass is = 9.75*10^-7 grams
1µg = 10^-6 g
You then have 9.75*10^-7/10^-6 = 0.975µg ZnC2O4
(*see below) at this point you could have said:
1µg = 10^-6 g therefore you have a solution of 6.29µg per litre,
155ml = 6.29*155/1000 = 0.975µg ZnC2O4</span>
C16H32O2(aq) --> 16CO2(g) + 16H2O(l) ... said its wrong though?
<span>This is because you haven't added any oxygen needed for the combustion, so your equation does'nt balance. Also a solution in water [aq] doesn't burn! </span>
<span>Try </span><span>C16H32O2(s) + 23O2(g) --> 16CO2(g) + 16H2O(l)
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