Correct answer is <span>Fuels do not have to be purchased to generate power.</span>
0.88qt
Explanation:
Given parameters:
Density of ebony wood = 1.174g/ml
Mass of wood = 2.1549lb
Unknown:
Volume of the wood in quarts = ?
Solution:
Density is defined as the mass per unit volume of a substance.
Mass is the quantity of matter the wood contains.
Volume is the space it occupies.
To solve this problem, convert the mass form pounds into grams and plug into the density equation to find volume.
Then convert the volume from mL to quarts.
Density = 
Volume = 
1 pound = 453.592g
2.1549 lb = 2.1549 x 453.592 = 977.5g
Volume =
= 832.6mL
1 quart = 946.353mL
Therefore 832.6mL x
= 0.88qt
learn more:
Density brainly.com/question/8441651
#learnwithBrainly
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.