Given:
M = 0.0150 mol/L HF solution
T = 26°C = 299.15 K
π = 0.449 atm
Required:
percent ionization
Solution:
First, we get the van't Hoff factor using this equation:
π = i MRT
0.449 atm = i (0.0150 mol/L) (0.08206 L atm / mol K) (299.15 K)
i = 1.219367
Next, calculate the concentration of the ions and the acid.
We let x = [H+] = [F-]
[HF] = 0.0150 - x
Adding all the concentration and equating to iM
x +x + 0.0150 - x = <span>1.219367 (0.0150)
x = 3.2905 x 10^-3
percent dissociation = (x/M) (100) = (3.2905 x 10-3/0.0150) (100) = 21.94%
Also,
percent dissociation = (i -1) (100) = (</span><span>1.219367 * 1) (100) = 21.94%</span>
Answer:
35
Explanation:
because it's is still the same amount even if you freeze it
Answer: There are
five <span>bonding pairs of electrons in Methanol.
Explanation: Those electron pairs which are being shared between two atoms in molecule are called as
bonding pair electrons. While, those electron pairs which are not involved in bond formation and are not shared between two atoms are called as
Non-Bonding electron pairs.
In molecule of
Methanol as shown below, it can be seen that carbon atom is forming four bonds with three hydrogen atoms and one oxygen atom by sharing four electron pairs and oxygen is forming two bonds, one with carbon atom and one with hydrogen atom. There are two lone pair of electrons present on oxygen atom which are not taking part in and formation.</span>
<span>According to the question-
1 mol C3H8O = 60.096 g C3H8O
2 mol C3H8O = 9 mol O2
1 mol O2 = 31.998 g O2
[(3.00 g C3H8O)/1][(1 mol C3H8O)/(60.096)][(9 mol O2)/(2 mol C3H8O)][(32.998 g O2)/(1 mol O2)] = 7.1880435 g O2
Since 7.1880435 g of O2 is needed, and 7.38 g of O2 is available, 0.199565 g of O2 will be left over and oxygen is present in excess.
Next, we need to convert 0.199565 g of O2 into moles of O2:
[(0.199565 g O2)/1][(1 mol O2)/(31.998 g O2)] = 0.005999 mol O2, or 0.006 mol O2</span>
The answer to this is true.