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.
Answer: 1) Temperature can change the solubility of a solute.
Explanation:
The chart is missing so there is no way to tell what does the graph show.
Yet, I can help you because I can explain the status of each statement of the choices. As you will see there is only one possibility..
<span>1) Temperature can change the solubility of a solute.
Yes, temperature definetly can, and mostly do, modify the solubility of a solute.
You can search any chart of solubility and will find that.
I can give you two examples:
a) Sodium chloride: dissolve some spoons of salt in a cold water until you can not dissolve more. Then, heat the water, you will find that more salt will get dissolved, proving that the temperature of the solution increases the solubility of sodium chloride.
b) Carbon dioxide gas: the soft drinks have CO₂ molecules dissolved in it.
The higher the temperature of the soft drink the less the amount of CO₂(g) that can be dissolved. That is why the soda bottling plants cool the beverage before adding the CO₂(g).
2) </span><span>Temperature has no affect on the solubility of a solute.
Since this is the opposite to the first statement and the first is true, this is false.
3) Salt has a greater solubility than sugar.
False.
This is an empirical result, which you cannot predict theoretically. So you need to see at the data either in a table or in a chart. Else you can test it at home. After the empirical data are shown it results that more grams of sugar can be dissolved in water compared to salt.
That is something you ca see in a chart or you can prove by yourself.
4) Nitrite salt has a greater solubility than sugar.
</span>
False.
Looking at some data you can find that sodium nitrite solutiliby is aroun 70 - 100 g/10 g while sugar (sucrose) solutiblity is around 180 - 235 g/ 100 g.
Answer
im not quite sure but I think the answer is <em>D atom</em><em> </em>
Explaination
Answer:
3000J
Explanation:
Formula: Work = Force × Distance
Solution: Work = 200N × 15m
Answer = 3000N/m or 3000J
Answer:
1) Heterogeneous mixture
2). homogeneous mixture
3) water is solvent and sugar is solute
4). sublimation process
