Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
Note: This question is incomplete and lacks very important data to solve this question. But I have found the similar question which shows the profiles about which question discusses. Using the data from that question, I have solved the question.
a) We need to find the major species from A to F.
Major Species at A:
1. 
Major Species at B:
1.
2. 
Major Species at C:
1.
Major Species at D:
1.
2. 
Major Species at E:
1.
Major Species at F:
1.
b) pH calculation:
At Halfway point B:
pH = pK
+ log[
]/[H
]
pH = pK = 6.35
Similarly, at halfway point D.
At point D,
pH = pK
+ log [H]/[H2]
pH = pK = 10.33
for 39g water solute dissolved at 20C = solubility ( g/ 100 g H2O ) × mass of water = ( 11g / 100g H2O ) × 39g H2O = 4.29 g
amount of solute dissolved at 30 C =
= 23 / 100 * 39 = 8.97 g
Amount of extra solute dissolved = 8.97 - 4.29 = 4.7 g
Answer:
false, Potassium and fluorine are not halogens.
only fluorine here is halogen.
potassium is an alkali earth metal it doesn't comes under category of halogens, but fluorine
is a non metal which comes under halogen family.
Answer:
0.500 mole of Xe (g) occupies 11.2 L at STP.
General Formulas and Concepts:
<u>Gas Laws</u>
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
<u>Stoichiometry</u>
- Mole ratio
- Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
<em>Identify.</em>
0.500 mole Xe (g)
<u>Step 2: Convert</u>
- [DA] Set up:
- [DA] Evaluate:
Topic: AP Chemistry
Unit: Stoichiometry
I think it would be Kriptonite
