how much hydrogen will be released during the interaction of potassium weighing 8 grams with hydrochloric acid
1 answer:
Answer:
With an understanding of the ideal gas laws, it is now possible to apply these principles to chemical stoichiometry problems. For example, zinc metal and hydrochloric acid (hydrogen chloride dissolved in water) react to form zinc (II) chloride and hydrogen gas according to the equation shown below:
2 HCl (aq) + Zn (s) → ZnCl2 (aq) + H2 (g)
Explanation:
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The provided information are:
volume of 0.85 M lactic acid = 225 ml
volume of 0.68 M sodium lactate = 435 ml
Ka of the lactate buffer = 1.38 x 10⁻⁴
The equation for dissociation of lactic acid is:
CH₃CH(OH)COOH(aq) + H₂O ⇄ CH₃CH(OH)COO⁻(aq) + H₃O⁺(aq)
The pH of buffer is calculated from Henderson-Hasselbalch equation, which is:
pH = pKa + log![\frac{[conjugated base]}{[Acid]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5Bconjugated%20base%5D%7D%7B%5BAcid%5D%7D%20)
pKa = - log Ka = - log (1.38 x 10⁻⁴) = 3.86
The number of moles of lactic acid and lactate are as follows:
n (Lactic acid) = 225 ml x 0.85 mmol/ml = 191.25 mmol
n (Lactate) = 435 ml x 0.68 mmol/ml = 295.8 mmol
The number of moles of lactic acid and lactate in total volume of the solution:
[CH₃CH(OH)COOH] = n (lactic acid) / 660 ml = 191.25 mmol / 660 ml = 0.29 M
[CH₃CH(OH)COO⁻] = n (lactate) / 660 ml = 0.45 M
pH = 3.86 + log
= 3.86 + 0.191 = 4.05
So the pH of given solution is 4.05
volume of 0.85 M lactic acid = 225 ml
volume of 0.68 M sodium lactate = 435 ml
Ka of the lactate buffer = 1.38 x 10⁻⁴
The equation for dissociation of lactic acid is:
CH₃CH(OH)COOH(aq) + H₂O ⇄ CH₃CH(OH)COO⁻(aq) + H₃O⁺(aq)
The pH of buffer is calculated from Henderson-Hasselbalch equation, which is:
pH = pKa + log
pKa = - log Ka = - log (1.38 x 10⁻⁴) = 3.86
The number of moles of lactic acid and lactate are as follows:
n (Lactic acid) = 225 ml x 0.85 mmol/ml = 191.25 mmol
n (Lactate) = 435 ml x 0.68 mmol/ml = 295.8 mmol
The number of moles of lactic acid and lactate in total volume of the solution:
[CH₃CH(OH)COOH] = n (lactic acid) / 660 ml = 191.25 mmol / 660 ml = 0.29 M
[CH₃CH(OH)COO⁻] = n (lactate) / 660 ml = 0.45 M
pH = 3.86 + log
So the pH of given solution is 4.05
Answer:
Density, melting point. and magnetic properties
Explanation:
I can think of three ways.
1. Density
The density of Cu₂S is 5.6 g/cm³; that of CuS is 4.76 g/cm³.
It should be possible to distinguish these even with high school equipment.
2. Melting point
Cu₂S melts at 1130 °C (yellowish-red); CuS decomposes at 500 °C (faint red).
A Bunsen burner can easily reach these temperatures.
3. Magnetic properties
You can use a Gouy balance to measure the magnetic susceptibilities.
In Cu₂S the Cu⁺ ion has a d¹⁰ electron configuration, so all the electrons are paired and the solid is diamagnetic.
In CuS the Cu²⁺ ion has a d⁹ electron configuration, so all there is an unpaired electron and the solid is paramagnetic.
A sample of Cu₂S will be repelled by the magnetic field and show a decrease in weight.
A sample of CuS will be attracted by the magnetic field and show an increase in weight.
In the picture below, you can see the sample partially suspended between the poles of an electromagnet.
