2KF(s) -----> 2K(s)+F2(g)
The initial concentration was 8.45M
Step 1: Given data
Initial concentration (C₁): ?
Initial volume (V₁): 635 mL = 0.845 L
Final concentration (C₂): 6.00 M
Final volume (V₂): 1.19 L
Step 2: Calculate the initial concentration
We have a concentrated NaCl solution and we want to prepare a diluted one. We will use the dilution rule.
C₁ × V₁ = C₂ × V₂
C₁ = C₂ × V₂ / V₁
C₁ = 6.00 M × 1.19 L / 0.845 L
C₁ = 8.45 M
Thus, The initial concentration was 8.45M
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Ok first, we have to create a balanced equation for the dissolution of nitrous acid.
HNO2 <-> H(+) + NO2(-)
Next, create an ICE table
HNO2 <--> H+ NO2-
[]i 0.230M 0M 0M
Δ[] -x +x +x
[]f 0.230-x x x
Then, using the concentration equation, you get
4.5x10^-4 = [H+][NO2-]/[HNO2]
4.5x10^-4 = x*x / .230 - x
However, because the Ka value for nitrous acid is lower than 10^-3, we can assume the amount it dissociates is negligable,
assume 0.230-x ≈ 0.230
4.5x10^-4 = x^2/0.230
Then, we solve for x by first multiplying both sides by 0.230 and then taking the square root of both sides.
We get the final concentrations of [H+] and [NO2-] to be x, which equals 0.01M.
Then to find percent dissociation, you do final concentration/initial concentration.
0.01M/0.230M = .0434 or
≈4.34% dissociation.
Answer:
No. 1 Susan was <u>Frantic</u>
No.2 panicked
No.3 determined
No.4 Despair
No.5 Releif
I really hope this helps, Good luck! :D
As the length of carbon chain increases the forces which hold the atoms also increases . Ethanol is a liquid due to higher bond energy
