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<u>Answer:</u> No crystals of potassium sulfate will be seen at 0°C for the given amount.
<u>Explanation:</u>
We are given:
Mass of potassium nitrate = 47.6 g
Mass of potassium sulfate = 8.4 g
Mass of water = 130. g
Solubility of potassium sulfate in water at 0°C = 7.4 g/100 g
This means that 7.4 grams of potassium sulfate is soluble in 100 grams of water
Applying unitary method:
In 100 grams of water, the amount of potassium sulfate dissolved is 7.4 grams
So, in 130 grams of water, the amount of potassium sulfate dissolved will be
As, the soluble amount is greater than the given amount of potassium sulfate
This means that, all of potassium sulfate will be dissolved.
Hence, no crystals of potassium sulfate will be seen at 0°C for the given amount.
P = 11.133 atm (purple)
T = -236.733 °C(yellow)
n = 0.174 mol(red)
<h3>Further explanation </h3>Some of the laws regarding gas, can apply to ideal gas (volume expansion does not occur when the gas is heated),:
- Boyle's law at constant T, P = 1 / V
- Charles's law, at constant P, V = T
- Avogadro's law, at constant P and T, V = n
So that the three laws can be combined into a single gas equation, the ideal gas equation
In general, the gas equation can be written
where
P = pressure, atm
V = volume, liter
n = number of moles
R = gas constant = 0.08206 L.atm / mol K
T = temperature, Kelvin
To choose the formula used, we refer to the data provided
Because the data provided are temperature, pressure, volume and moles, than we use the formula PV = nRT
- Purple box
T= 10 +273.15 = 373.15 K
V=5.5 L
n=2 mol
- Yellow box
V=8.3 L
P=1.8 atm
n=5 mol
- Red box
T = 12 + 273.15 = 285.15 K
V=3.4 L
P=1.2 atm