Answer:
pH of buffer is 4.81
Explanation:
The reaction between HCl and sodium acetate will product acetic acid. The presence of weak acid (acetic acid) and its salt will make a buffer.
The reaction will be:
Thus one mole each of HCl and acetate will give one mole of acetic acid.
The moles of HCl present = molarity X volume = 0.452 X 0.204 =0.092
moles of acetic acid formed on reaction of HCl with acetate = 0.092
Initial moles of sodium acetate present = molarity X volume = 0.4X0.5 =0.2
moles of sodium acetate left =0.108 mol
the pKa of acetic acid =4.74
The pH of buffer is calculated using Henderson Hassalbalch's equation :
pH = pKa + log
<u>Answer:</u> The solubility of in water is
<u>Explanation:</u>
The balanced equilibrium reaction for the ionization of cadmium phosphate follows:
3s 2s
The expression for solubility constant for this reaction will be:
We are given:
Putting values in above equation, we get:
Hence, the solubility of in water is
The integrated rate law for a second-order reaction is given by:
where, [A]t= the concentration of A at time t,
[A]0= the concentration of A at time t=0
<span>k =</span> the rate constant for the reaction
<u>Given</u>: [A]0= 4 M, k = 0.0265 m–1min–1 and t = 180.0 min
Hence,
<span> = 4.858</span>
<span><span><span>Therefore, [A]</span>t</span>= 0.2058 M.</span>
<span>
</span>
<span>Answer: C</span>oncentration of A, after 180 min, is 0.2058 M
Answer:
= 19
ΔG° of the reaction forming glucose 6-phosphate = -7295.06 J
ΔG° of the reaction under cellular conditions = 10817.46 J
Explanation:
Glucose 1-phosphate ⇄ Glucose 6-phosphate
Given that: at equilibrium, 95% glucose 6-phospate is present, that implies that we 5% for glucose 1-phosphate
So, the equilibrium constant can be calculated as:
= 19
The formula for calculating ΔG° is shown below as:
ΔG° = - RTinK
ΔG° = - (8.314 Jmol⁻¹ k⁻¹ × 298 k × 1n(19))
ΔG° = 7295.05957 J
ΔG°≅ - 7295.06 J
b)
Given that; the concentration for glucose 1-phosphate = 1.090 x 10⁻² M
the concentration of glucose 6-phosphate is 1.395 x 10⁻⁴ M
Equilibrium constant can be calculated as:
0.01279816514 M
0.0127 M
ΔG° = - RTinK
ΔG° = -(8.314*298*In(0.0127)
ΔG° = 10817.45913 J
ΔG° = 10817.46 J