Answer:
balanced
Explanation:
its balance since all the elments add up on both sides
when the thermal energy is the energy contained within a system that is responsible for its temperature.
and when the thermal energy is can be determined by this formula:
q = M * C *ΔT
when q is the thermal energy
and M is the mass of water = 100 g
and C is the specific heat capacity of water = 4.18 joules/gram.°C
and T is the difference in Temperature = 50 °C
So by substitution:
∴ q = 100 g * 4.18 J/g.°C * 50
= 20900 J = 20.9 KJ
If the conjugate base of a molecule has a pKb of 1.4, the molecule should be a Weak Acid.
Notice this question gives us the pKb of the molecule, not the pKa. Because of this, the pH scale basically gets reversed, so lower numbers in pKb correlate with stronger bases, and higher numbers in pKb correlate with stronger acids - the exact opposite of the pH scale.
It's important to make sure you completely understand the terms of conjugate base, conjugate acid, pKb, pKa, and how they all relate. It's easy to mix up the meanings of these definitions.
Here are the two other pieces of information you need to know to correctly answer this question:
- Strong acids have a weak conjugate base.
- Strong bases have a weak conjugate acid.
So if the problem says you have a strong conjugate base, then the molecule must be a weak acid. To illustrate this, think of ammonium, NH4+. Ammonium is a weak acid, but the conjugate base of ammonium is ammonia, NH3, which is a reasonably good base.
Learn more about conjugate base here : brainly.com/question/22514615
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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:
520ML and apparently I need to put more in this answer
Explanation:
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