They try to base their conclusions off of data and measurements of which they should record from conducting experiments
<u>Answer:</u> The value of <em>i</em> is 1.4 and 40% dissociation of 100 particles of zinc sulfate will yield 60 undissociated particles.
<u>Explanation:</u>
The equation used to calculate the Vant' Hoff factor in dissociation follows:
where,
= degree of dissociation = 40% = 0.40
i = Vant' Hoff factor
n = number of ions dissociated = 2
Putting values in above equation, we get:
The equation used to calculate the degee of dissociation follows:
Total number of particles taken = 100
Degree of dissociation = 40% = 0.40
Putting values in above equation, we get:
This means that 40 particles are dissociated and 60 particles remain undissociated in the solution.
Hence, 40% dissociation of 100 particles of zinc sulfate will yield 60 undissociated particles.
The answer would be .5 mols because you take the total amount of grams, which is 20, and you had up the molar mass of sodium hydroxide, which would be 40. After you have this you would set this up as a stochiometry equation. With 1 mol on top you dived 20/40 to cancel out your grams. This leaves you with .5 mols
Answer:
The specific heat of zinc is 0.361 J/g°C
Explanation:
<u>Step 1:</u> Data given
44.0 J needed
Mass of solid zinc = 10.6 grams
Initial temperature = 24.9 °C
Final temperature = 36.4 °C
<u>Step 2</u>: Calculate the specific heat of zinc
Q = m*c*ΔT
⇒ with Q = heat (in Joule) = 44.0 J
⇒ with m = the mass of the solid zinc = 10.6 grams
⇒ with c = the specific heat of the zinc = TO BE DETERMINED
⇒ with ΔT = The change in temperature = T2-T1 = 36.4 °C - 24.9 °C = 11.5 °C
44.0 J = 10.6 grams * c * 11.5°C
c = 44.0 J / (10.6g * 11.5 °C)
c = 0.361 J/g°C
The specific heat of zinc is 0.361 J/g°C
<u>Given information:</u>
Concentration of NaF = 0.10 M
Ka of HF = 6.8*10⁻⁴
<u>To determine:</u>
pH of 0.1 M NaF
<u>Explanation:</u>
NaF (aq) ↔ Na+ (aq) + F-(aq)
[Na+] = [F-] = 0.10 M
F- will then react with water in the solution as follows:
F- + H2O ↔ HF + OH-
Kb = [OH-][HF]/[F-]
Kw/Ka = [OH-][HF]/[F-]
At equilibrium: [OH-]=[HF] = x and [F-] = 0.1 - x
10⁻¹⁴/6.8*10⁻⁴ = x²/0.1-x
x = [OH-] = 1.21*10⁻⁶ M
pOH = -log[OH-] = -log[1.21*10⁻⁶] = 5.92
pH = 14 - pOH = 14-5.92 = 8.08
Ans: (b)
pH of 0.10 M NaF is 8.08
