Answer:
The mol fraction of sucrose in an aqueous solution at 20°C is 0.114
Explanation:
Step 1: Data given
The vapor pressure of water above the solution is 2.0 mmHg
The vapor pressure of pure water at 20 °C is 17.5 mmHg
Step 2: Calculate the mol fraction of sucrose
Psolution = (χsolvent) (P°solvent)
⇒with Psolution = the vapor pressure of the solution is 2.0 mmHg
⇒with χsolvent = the mol fraction of sucrose = TO BE DETERMINED
⇒with P°solvent = the vapor pressure of pure water at 20 °C =17.5 mmHg
2.0 mmHg = X * 17.5 mmHg
X = 2.0 mmHg / 17.5 mmHg
X = 0.114
The mol fraction of sucrose in an aqueous solution at 20°C is 0.114
In oxygen- starved muscle cells in anaerobic respiration lactic acid is formed while in anaerobic respiration by yeast cells it produces ethanol and carbon dioxide.
Explanation:
Respiration occurring in absence of oxygen in muscles leads to accumulation of lactic acid and very less amount of energy. This is fermentation process as the end product formed is lactic acid.
The process of fermentation takes place in yeast bacteria and oxygen starved cells of muscles.
The yeast can ferment simple sugars like glucose and fructose into ethyl alcohol and carbon dioxide.
In general human respiration is aerobic but during strenuous exercise, oxygen does not meet the demand of working out muscle, so they respire anaerobically and produce lactic acid and 2 ATP i.e very less compared to 38 molecules of ATP in aerobic respiration.
Answer:
Concentration: 0.185M HX
Ka = 9.836x10⁻⁶
pKa = 5.01
Explanation:
A weak acid, HX, reacts with NaOH as follows:
HX + NaOH → NaX + H2O
<em>Where 1 mole of HX reacts with 1 mole of NaOH</em>
To solve this question we need to find the moles of NaOH at equivalence point (Were moles HX = Moles NaOH).
18.50mL = 0.01850L * (0.20mol / L) = 0.00370 moles NaOH = Moles HX
In 20.0mL = 0.0200L =
0.00370 moles HX / 0.0200L = 0.185M HX
The equilibrium of HX is:
HX(aq) ⇄ H⁺(aq) + X⁻(aq)
And Ka is defined as:
Ka = [H⁺] [X⁻] / [HX]
<em>Where [H⁺] = [X⁻] because comes from the same equilibrium</em>
As pH = 2.87, [H+] = 10^-pH = 1.349x10⁻³M
Replacing:
Ka = [H⁺] [H⁺] / [HX]
Ka = [1.349x10⁻³M]² / [0.185M]
Ka = 9.836x10⁻⁶
pKa = -log Ka
<h3>pKa = 5.01</h3>
Can still be used to do work
