Answer:
The ΔH of the reaction is + 12.45 KJ/mol
Explanation:
Mass of water= 100ml = 100g. (You should always assume 1cm3 of water as 1g)
heat capacity of water = 4.18 Jk-1 Mol-1
Change in temperature = (19.86 - 25.00) = -5.14 K (This is an endothermic reaction because of the fall in temperature)
Molar mass of NaHCO3 = 84 g/mol
Mole of NaHCO3 = 14.5 / 84 = 0.173 mol
Step 1 : Calculate the heat energy (Q) lost by the water.
Q = M x C x ΔT
Q = -100 x 4.18 x (-5.14)
Q = 2148.5 joules
Q = 2.1485 K J
Step 2: Calculating the ΔH of the reaction?
ΔH = Q / number of moles of NaHCO3
ΔH = 2.1485 / 0.173
ΔH = 12.42 KJ/mol
Set a piece of pH paper in front of each solution before testing to keep organized.
Use a transfer pipet to remove a few drops from the solution to drop onto the paper.
Wait a minute or so before reading color.
Is the correct answer...
Answer:
Option A; V = 2.92 L
Explanation:
If we assume a lot of things, like:
The gas is an ideal gas.
The temperature is constant.
The gas does not interchange mass with the environment.
Then we have the relation:
P*V = n*R*T = constant.
Where:
P = pressure
V = volume
n = number of moles
R = constant of the ideal gas
T = temperature.
We know that when P = 0.55 atm, the volume is 5.31 L
Then:
(0.55 atm)*(5.31 L) = constant
Now, when the gas is at standard pressure ( P = 1 atm)
We still have the relation:
P*V = constant = (0.55 atm)*(5.31 L)
(1 atm)*V = (0.55 atm)*(5.31 L)
Now we only need to solve this for V.
V = (0.55 atm/ 1 atm)*(5.31 L) = 2.92 L
V = 2.92 L
Then the correct option is A.