An ionic bond is composed of a cation and anion, which are charged ions. You cannot have atoms with ionic bonds, because there is no opposite attraction between charges due to atoms having a net nuclear charge of zero. Therefore, a cation and anion would most likely be joined by an ionic bond.
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:
H+(aq) + OH-(aq) → H2O(l)
Explanation:
Step 1: Data given
nitrious acid = HNO3
sodium hydroxide = NaOH
Step 2: The unbalance equation
HNO3(aq) + NaOH(aq) →NaNO3(aq) + H2O(l)
The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side (Ba^2+ and Br-), look like this:
H+(aq) + NO3-(aq) + Na+(aq) + OH-(aq) →Na+(aq) +NO3(aq) + H2O(l)
H+(aq) + OH-(aq) → H2O(l)
Answer:
Part A: 47.8 mi/h
Part B: 0.072 M/s
Part C: 0.144 M/s
Explanation:
Part A
The average speed or velocity (V) is the variation of the space divided by the variation of the time:
V = (241 - 2)/(8 -3)
V = 47.8 mi/h
Part B
As Part A, the average rate (r) of formation of I2 is the variation of the concentration divided by the variation of time:
r = (1.83 - 1.11)/(15 - 5)
r = 0.072 M/s
Part C
The rates of the substances are proportional of their number of moles (n) which are their coefficient, so:
rI2/nI2 = rHCl/nHCl
0.072/1 = rHCl/2
rHCl = 2*0.072
rHCl = 0.144 M/s