The first dissociation for H2X:
H2X +H2O ↔ HX + H3O
initial 0.15 0 0
change -X +X +X
at equlibrium 0.15-X X X
because Ka1 is small we can assume neglect x in H2X concentration
Ka1 = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15)
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
HX + H2O↔ X^2 + H3O
8.2x10^-4 Y 8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2 = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m
Answer:
A campfire is a perfect example of the different kinds of heat transfer. If you boil water in a kettle, the heat is transferred through convection from the fire to the pot.
Explanation:
True because it actually formed I know
Answer:
The van't Hoff factor is 2.55
Explanation:
Step 1: Data given
Osmotic pressure of CaCl2 = 0.585 atm
Osmotic pressure of urea = 0.237 atm
Concentration = 9.69 * 10^-3 M
Temperature = 25.0 °C
Step 2:
Π = iMRT
⇒ with Π = the osmotic pressure of CaCl2 = 0.585 atm
⇒ with i is the van't Hoff factor
⇒ with M = the molar concentration = 9.69 * 10^-3 M
⇒ with R = the gas constant = 0.0821 L*atm/mol*K
⇒ with T = the temperature = 25.0 °C = 298 K
0.605 atm = i(9.69 * 10^-3 M)(0.0821 L*atm/mol*K)(298K)
i = 2.55
The van't Hoff factor is 2.55
Calcium with water:
When a metal reacts with water( cold water or hot water) then the products formed are metal hydroxide and hydrogen gas.
Metal + steam --------> Metal hydroxide + Hydrogen
Calcium reacts with cold water to form calcium hydroxide and hydrogen gas:
potassium with water.
Potassium reacts violently with cold water to form potassium hydroxide and hydrogen gas:
2K(s)+ 2H2O(l) --------> 2KOH(aq) + H2(g) + heat.
In this reaction so much heat is produced get hydrogen gas formed catches fire and burns explosively.