Answer:
- About 18 g of NH₄Cl will precipitate.
Explanation:
The <em>table G</em> is the graph of the solubility curves for several solutes which is attached.
The second picture identifies the solubilities for the NH₄Cl at 50ºC and 10ºC.
The solubility of NH₄Cl at 50ºC is about 52 g/ 100 g of water.
The solubility of NH₄Cl at 10ºC is about 34 g / 100 g of water.
Then, at 50ºC 100 g of water saturated with NH₄Cl contains about 52 g of NH₄Cl and 100 g of water saturated with NH₄Cl contains 34 g of NH₄Cl.
The difference, 52g - 34 g of NH₄Cl shall precipitate:
52 g - 34 g = 18g ← answer
This says four figures. The 4 figures you should use are 4546 with a peek at 7 to see what effect it will have on the 4 main figures.
45.467 rounds to 45.47
You round the 4th figure up one. You are not concerned about the decimal places in this question.
I am almost positive that the answer is D
Answer:
4.81 moles
Explanation:
The total pressure of the gas = Pressure at which gauge reads zero + pressure read by it.
Pressure at which gauge reads zero = 14.7 psi
Pressure read by the gauge = 988 psi
Total pressure = 14.7 + 988 psi = 1002.7 psi
Also, P (psi) = P (atm) / 14.696
Pressure = 1002.7 / 14.696 = 68.2297 atm
Temperature = 25 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (25 + 273.15) K = 298.15 K
Volume = 1.50 L
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
68.2297 atm × 1.5 L = n × 0.0821 L.atm/K.mol × 298.15 K
⇒n = 4.81 moles