Sugar is a compound.
<span>C12H22O<span>11</span></span>
Answer:
Explanation:
Because 3.005 grams of potassium lactate is added to 100. mL of solution, its concentration is:
![\displaystyle \begin{aligned} \left[ \text{KC$_3$H_$_5$O$_3$}\right] & = \frac{3.005\text{ g KC$_3$H_$_5$O$_3$}}{100.\text{ mL}} \cdot \frac{1\text{ mol KC$_3$H_$_5$O$_3$}}{128.17 \text{ g KC$_3$H_$_5$O$_3$}} \cdot \frac{1000\text{ mL}}{1\text{ L}} \\ \\ &= 0.234\text{ M}\end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20%5Cleft%5B%20%5Ctext%7BKC%24_3%24H_%24_5%24O%24_3%24%7D%5Cright%5D%20%20%26%20%3D%20%5Cfrac%7B3.005%5Ctext%7B%20g%20KC%24_3%24H_%24_5%24O%24_3%24%7D%7D%7B100.%5Ctext%7B%20mL%7D%7D%20%5Ccdot%20%5Cfrac%7B1%5Ctext%7B%20mol%20KC%24_3%24H_%24_5%24O%24_3%24%7D%7D%7B128.17%20%5Ctext%7B%20g%20KC%24_3%24H_%24_5%24O%24_3%24%7D%7D%20%5Ccdot%20%5Cfrac%7B1000%5Ctext%7B%20mL%7D%7D%7B1%5Ctext%7B%20L%7D%7D%20%5C%5C%20%5C%5C%20%26%3D%200.234%5Ctext%7B%20M%7D%5Cend%7Baligned%7D)
By solubility rules, potassium is completely soluble, so the compound will dissociate completely into potassium and lactate ions. Therefore, [KC₃H₅O₃] = [C₃H₅O₃⁺]. Note that lactate is the conjugate base of lactic acid.
Recall the Henderson-Hasselbalch equation:
![\displaystyle \begin{aligned}\text{pH} = \text{p}K_a + \log \frac{\left[\text{Base}\right]}{\left[\text{Acid}\right]} \end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%5Ctext%7BpH%7D%20%3D%20%5Ctext%7Bp%7DK_a%20%2B%20%5Clog%20%5Cfrac%7B%5Cleft%5B%5Ctext%7BBase%7D%5Cright%5D%7D%7B%5Cleft%5B%5Ctext%7BAcid%7D%5Cright%5D%7D%20%5Cend%7Baligned%7D)
[Base] = 0.234 M and [Acid] = 0.500 M. We are given that the resulting pH is 3.526. Substitute and solve for p<em>Kₐ</em>:
In conclusion, the p<em>Kₐ </em>value of lactic acid is about 3.856.
<span>The answer is 200 mol of water.
The balanced reaction is 2(H2) + (O2) = 2(H2O)
The limiting reactant is O2 as it will be completely consumed first, before hydrogen gas. Hydrogen gas would need at least 105 mol oxygen gas to be consumed; in excess of the 100 mol O2.
Looking at the stoichiometric coefficients, the ratio between water and oxygen is 2:1.
Therefore, the water produced would be 200 moles.</span>
Answer:
1.99 M
Explanation:
The molar mass of sodium thiosulfate (solute) is 158.11 g/mol. The moles corresponding to 110 grams are:
110 g × (1 mol/158.11 g) = 0.696 mol
The volume of solution is 350 mL = 0.350 L.
The molarity of sodium thiosulfate is:
M = moles of solute / liters of solution
M = 0.696 mol / 0.350 L
M = 1.99 M
Answer:
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Explanation:
Approximately 2 mL of Solution A (on the left) is added to a sample of Solution B (on the right) with a dropping pipet. If a precipitate forms, the resulting precipitate is suspended in the mixture. The mixture is then stirred with a glass stirring rod and the precipitate is allowed to settle for about a minute.
Solution A: 0.5 M sodium hydroxide, colorless
Solution B: 0.2 M nickel(II) nitrate, green
Precipitate: light green
Ni(NO3)2(aq) + 2 NaOH(aq) —> Ni(OH)2(s) + 2 NaNO3(aq)
Credits:
Design
Kenneth R. Magnell Central Michigan University, Mt. Pleasant, MI 48859
John W. Moore University of Wisconsin - Madison, Madison, WI 53706
Video
Jerrold J. Jacobsen University of Wisconsin - Madison, Madison, WI 53706
Text
Kenneth R. Magnell Central Michigan University, Mt. Pleasant, MI 48859
