Answer: Photosynthetic organisms take light energy and use it to make their own "food". In this process they use carbon dioxide and light to make sugars and oxygen. This is great for other animals that utilize this oxygen and eat the plants for food.
HOPE THIS HELPS
Missing in your question :
Concentration by (M):
Xy: y: X
initial 0.2 0.3 0.3
change +X -X -X
equilibrim (0.2+x) (0.3-x) (0.3-x)
according to Kc formula: when Kc = 0.2
Kc = [XY]/[X]*[Y]
0.2 = (0.2+x) / (0.3-x)*(0.3-x)
0.2=(0.2+x) / (0.3-x)^2 by solving this equation
0.2*(0.3-x)^2 = 0.2+x
0.2* (0.09-0.6x+x^2)= 0.2 +x
0.0018 - 0.12 X +0.2X^2 = 0.2 + X
0.2X^2 -1.12 X -0.1982 = 0
∴X= 0.17
∴[XY] = 0.2 + 0.17 = 0.37 m
∴[X] = 0.3 - 0.17 =0.13 m
∴[y] = 0.3 - 0.17 = 0.13 m
The triprotic acid like H₃PO₄ contains three protons H⁺ so it ionized through three steps, the acidity strength of first proton higher than second proton higher than third one so Ka1 > Ka2 > Ka3 But
For pKa which equals to -log Ka ... the higher the value of Ka, the smaller the value of pKa. so the correct answer will be:
pKa1 < pKa2 < pKa3
Answer:
moles of camphor = 0.0522 moles
Explanation:
<u>Data:</u>
MW camphor= 152 g/mol
V solution = 1.32 L
M solution = 6.01 M
moles solute = ?
-
To calculate the moles of camphor, you must first know the grams of solute (camphor) that exist in the solution, this is calculated from the molarity equation:

-
From there the grams of the solute (camphor) are cleared:

-
Then by means of the molecular weight (MW) equation the moles can be obtained:


Answer:
1.27 atm is the final pressure of the oxygen in the flask (with the stopcock closed).
2.6592 grams of oxygen remain in the flask.
Explanation:
Volume of the flask remains constant = V = 2.0 L
Initial pressure of the oxygen gas = 
Initial temperature of the oxygen gas = 
Final pressure of the oxygen gas = 
Final temperature of the oxygen gas = 
Using Gay Lussac's law:


1.27 atm is the final pressure of the oxygen in the flask (with the stopcock closed).
Moles of oxygen gas = n
(ideal gas equation)

Mass of 0.08310 moles of oxygen gas:
0.08310 mol × 32 g/mol = 2.6592 g
2.6592 grams of oxygen remain in the flask.