The compound IF5 contains Question 16 options: polar covalent bonds with partial negative charges on the F atoms. ionic bonds. p
1 answer:
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So let's convert this amount of mL to grams:
Then we need to convert to moles using the molar weight found on the periodic table for mercury (Hg):
Then we need to convert moles to atoms using Avogadro's number:
So now we know that in 1.2 mL of liquid mercury, there are present.
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:
4.26 %
Explanation:
There is some info missing. I think this is the original question.
<em>Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10 ⁻⁴.</em>
<em />
Step 1: Given data
Initial concentration of the acid (Ca): 0.249 M
Acid dissociation constant (Ka): 4.50 × 10 ⁻⁴
Step 2: Write the ionization reaction for nitrous acid
HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)
Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])
We will use the following expression.
Step 4: Calculate the percent ionization of nitrous acid
We will use the following expression.
The pressure of the gas in the flask (in atm) when Δh = 5.89 cm is 1.04 atm
<h3>Data obtained from the question</h3>The following data were obtained from the question:
- Atmospheric pressure (Pa) = 730.1 torr = 730.1 mmHg
- Change in height (Δh) = 5.89 cm
- Pressure due to Δh (PΔh) = 5.89 cmHg = 5.89 × 10 = 58.9 mmHg
- Pressure of gas (P) =?
The pressure of the gas can be obtained as illustrated below:
P = Pa + PΔh
P = 730.1 + 58.9
P = 789 mmHg
Divide by 760 to express in atm
P = 789 / 760
P = 1.04 atm
Thus, the pressure of the gas when Δh = 5.89 cm is 1.04 atm
Learn more about pressure:
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Missing part of question:
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