N₂ + 3H₂ ⇒ 2NH₃
1mol : 2mol
3,72mol : 7,44mol
n = 7,44mol
M = 17g/mol
m = n * M = 7,44mol * 17g/mol = 126,48g
Boiling-point is the point of a pure liquid matter starts to evaporate and change into gaseous phase. It is where the set of conditions such as the pressure and temperature enough to do so. Boiling-point elevation, on the other hand, is the phenomenon of which the boiling point of a pure liquid matter is elevated because of the dissolved substances. A great example would be the boiling point of a distilled water (pure water) which is lesser than the boiling point of a sea water because of the dissolved salts. A pure water boils at 100°C at atmospheric pressure while a salt water boils at higher temperature than 100°C at the same pressure. Thus, the answer is D.
An aqueous solution in a 55 gallon (208 l drum), characterized by minimal buffering capacity, received 4kg of phenol and 1.5 kg of sodium phenate. What is the ph of the solution. The pka of phenol = 9.98. Mw of phenol and sodium phenate are 94 g/mol and 116 g/mol, respectively.
Volume of solution = 55 gallons = 208.2 L [ 1 gallon = 3.78 L]
moles of phenol = mass / molar mass = 4000 g / 94 = 42.55 moles
moles of sodium phenate = mass / molar mass = 1500 / 116 = 12.93 moles
pKa of phenol = 9.98
We know that the pH of buffer is calculated using Hendersen Hassalbalch's equation
pH = pKa + log [salt] / [acid]
volume is same for both the sodium phenate and phenol has we can directly take the moles of each in the formula
pH = 9.98 + log [12.93 / 42.55] = 9.46
4p, 3d, 3p, 3s, 2p, 2s and 1s orbitals may be occupied during de-excitation.<span />
The solubility of a sample will DECREASE when the size of the sample increases.
The bigger a substance is, the more will be the particles that make up this substance and the greater the amount of solvent that will be needed to dissolve the substance. Surface area of the substance is also important, a small surface area will impede solubility. Thus, when the size of a sample increases, the solubility decreases.
