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Answer:
The partial pressure of ammonia at equilibrium when a sufficient quantity of ammonium iodide is heated to 400°C Is 0.103 atm.
The correct option is A.
Explanation;
NH4I(s) ⇋ NH3(g) + HI(g)Kp = 0.215 at 400°C
NH4I(s)= 0.215
NH3(g)=0.103
HI(g)Kp=0.112
Therefore = 0.103 +0.112= 0.215
Therefore the partial pressure of ammonia at equilibrium is 0.103 atm
Answer:
what to do in this question
Answer:
Flash point of Lube Oil is around the 187°C mark,
Flash point of Biodiesel of 130°C
Flash point of Diesel Ranging from 52° to 96°
Answer:
15.4 g of sucrose
Explanation:
Formula to be applied for solving these question: colligative property of freezing point depression. → ΔT = Kf . m
ΔT = Freezing T° of pure solvent - Freezing T° of solution
Let's replace data given: 0°C - (-0.56°C) = 1.86 C/m . m
0.56°C / 1.86 m/°C = m → 0.301 mol/kg
m → molality (moles of solute in 1kg of solvent)
Our mass of solvent is not 1kg, it is 150 g. Let's convert it from g to kg, to determine the moles of solute: 150 g. 1kg/1000g = 0.150 kg
0.301 mol/kg . 0.150kg = 0.045 moles.
We determine the mass of sucrose, by the molar mass:
0.045 mol . 342 g/1mol = 15.4 g
