Answer:
91.7 kJ
Explanation:
Step 1: Given data
- Mass of ammonia (m): 66.7 g
- Molar heat of vaporization of ammonia (ΔH°vap): 23.4 kJ/mol
Step 2: Calculate the moles (n) corresponding to 66.7 g of ammonia
The molar mass of ammonia is 17.03 g/mol.
66.7 g × 1 mol/17.03 g = 3.92 mol
Step 3: Calculate the heat (Q) required to boil 3.92 moles of ammonia
We will use the following expression.
Q = ΔH°vap × n
Q = 23.4 kJ/mol × 3.92 mol = 91.7 kJ
Answer:
. ¿Qué
Explanation:
Hope you have a great day
Answer:
The compound you will use is the Dibasic phosphate
Explanation:
Simple phosphate buffer is used ubiquitously in biological experiments, as it can be adapted to a variety of pH levels, including isotonic. This wide range is due to phosphoric acid having 3 dissociation constants, (known in chemistry as a triprotic acid) allowing for formulation of buffers near each of the pH levels of 2.15, 6.86, or 12.32. Phosphate buffer is highly water soluble and has a high buffering capacity,
In this case the most efficient way is to disolve the dibasic compound which in the reaction with the water will form the monobasic phosphate.
To make the buffer you have to prepare the amount of distillate water needed, disolve the dibasic phospate, and then adjust with HCl or NaOH depending on the pH needed.
To answer this question, we will use the general gas law which states that:
PV = nRT where:
P is the pressure of the gas = <span>10130.0 kPa
</span>V is the volume of the gas = 50 liters
n is the number of moles that we want to calculate
R is the gas constant = <span>8.314 L∙kPa/K∙mol
T is the temperature = 300+273 = 573 degree kelvin
Substitute with the givens in the equation to get the number of moles as follows:
</span><span>10130 * 50 = n * 8.314 * 573
506500 = 4763.922 n
n = </span>506500 / 4763.922
n = 106.3199 moles
