Ok thanks for the valuble info.
Answer:
All chemical equations goes to the law of conservation of mass which says that matter can not be destroyed nor created which means there has to be an equal amount of atoms of each element on both sides of the equation. I hope that helps, I'm learning this as well.
The amount of substance present in a certain object with a given half-life in terms of h can be expressed through the equation,
A(t) = (A(o))(0.5)^(t/h)
where A(t) is the amount of substance after t years and A(o) is the original amount. In this item we are given that A(t)/A(o) is equal to 0.89. Substituting the known values,
0.89 = (0.5)(t / 5730 years)
The value of t from the equation is 963.34 years.
<em>Answer: 963 years</em>
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.
I think the answer is -12.7
