Explanation:
(a) The given data is as follows.
mass = 1 kg = 1000 g (as 1 kg = 1000 g)
Molar mass of 
= 17 g/mol
= 2.5 bar = 
(as 1 bar = 
)
= 5 bar = 
= 30 + 273 = 303 K
For adiabatic process, 
= constant = k
= 1.33 = 
(as PV = nRT)
= 
= 0.352 
Also, w = 
= 
= -84318.2 J
As 1 kJ = 1000 J. So, -84318.2 J = -84.318 kJ
Hence, the work required in kJ is -84318.2 J.
(b) It is known that for adiabatic system Q = 0,
= -w
dT = 
= 
= 895.93 K
We known that dT = 
so, 895.93 = 303 K - 
= (895.93 - 303)K
= 592.93 K
= (592.93 - 273.15)^{o}C
= 
Hence, the final temperature is .
Answer:
146g
Explanation:
We're assuming the reaction is occurring with excess chloride. First convert hydrogen mass to moles using molar mass - you need to work with numbers for these sorts of questions:
n = m / MM
n = 2g / 1g/mol
n = 2mol
Now use your stoich ratio to find how many moles of HCL are produced. There are 2 moles of HCl produced for every 1 mole of H2 reacted so:
x HCl / 2 mol H2 = 2 HCl / 1 H2
x = 4mol HCl
Now convert this back into mass using the molar mass of HCl:
m = n x MM
m = 4mol x 36.5g/mol
m = 146g
Answer:
-1.82 °C
Explanation:
Step 1: Given data
- Mass of NaCl (solute): 33.9 g
- Mass of water (solvent): 578 g = 0.578 kg
- Freezing point depression constant for water (Kb): -1.82 °C/m
Step 2: Calculate the molality of the solution
We will use the following expression.
m = mass of solute / molar mass of solute × kg of solvent
m = 33.9 g / 58.44 g/mol × 0.578 kg
m = 1.00 m
Step 3: Calculate the freezing point depression (ΔT)
The freezing point depression is a colligative property that, for a non-dissociated solute, can be calculated using the following expression:
ΔT = Kb × m
ΔT = -1.82 °C/m × 1.00 m
ΔT = -1.82 °C
It is called a proton I hope this helps
