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.
Answer:
Mass of the salt: 105.6g of KCl.
Mass water: 958.9g of water.
Molality: 1.478m.
Explanation:
<em>Mass of the salt:</em>
In 1L, there are 1.417 moles. In grams:
1.417 moles KCl * (74.54g / mol) = 105.6g of KCl
<em>Mass of the water:</em>
We can determine the mass of solution (Mass of water + mass KCl) by multiplication of the voluome (1L and density 1064.5g/L), thus:
1L * (1064.5g / L) = 1064.5g - Mass solution.
Mass water = 1064.5g - 105.6g = 958.9g of water
<em>Molality:</em>
Moles KCl = 1.417 moles KCl.
kg Water = 958.9g = 0.9589kg.
Molality = 1.417mol / 0.9589kg = 1.478m
Answer is: <span>because dissolved compounds can crystallizing from solution during filtration and forming crystals on the filter paper or funnel.
</span>Recrystallization<span> is a technique used to purify chemicals by dissolving both impurities and a compound in an appropriate solvent, either compound or impurities can be removed from the solution, leaving the other behind.</span>
You'll want to add three amounts of heat.
(1) Specific heat of lowering the temperature from -135°C to the melting point -114°C
(2) Latent heat of fusion/melting
(3) Specific heat of elevating the temperature from -114°C to -50°C
(1) E = mCΔT = (25 g)(0.97 J/g·°C)(1 kJ/1000 J)(-114 - -135) = 0.509 kJ
(2) E = mΔH = (25 g)(5.02 kJ/mol)(1 mol/46.07 g ethanol) = 2.724 kJ
(3) E = mCΔT = (25 g)(2.3 J/g·°C)(1 kJ/1000 J)(-50 - -114) = 3.68 kJ
<em>Summing up all energies, the answer is 6.913 kJ.</em>
