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stira [4]
3 years ago
5

What minimum concentration of clâ is required to begin to precipitate pbcl2? for pbcl2, ksp=1.17Ã10â5?

Chemistry
1 answer:
Juliette [100K]3 years ago
5 0
Missing in your question : [Pb2+] = 0.085 M

So from this balanced equation:
PbCl2  ↔ Pb+2   +   2Cl-

when Ksp = [Pb+2][Cl]^2
when we have Ksp = 1.17x10^-5 & [Pb+2] = 0.085 M so by substitution:
1.17x10^-5 = 0.085 [Cl-]^2

∴[Cl]^2 = (1.17x10^-5) / 0.085 = 1.38x10^-4
∴[Cl-] = √0.0002
          = 0.0117 M
So 0.0117 M is the minimum concentration requires precipitating PbCl2
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Calculate the standard potential for the following galvanic cell: Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s) which has the overall bala
mylen [45]

Answer:

1.06  V  

Explanation:

The standard reduction potentials are:

Ag^+/Ag     E° =  0.7996 V  

Ni^2+/Ni     E° = -0.257   V

The half-cell and cell reactions for Ni | Ni^2+ || Ag^+ | Ag are

Ni → Ni^2+ + 2e-                     E° = 0.257   V

<u>2Ag^+ 2e- → 2Ag               </u>    <u>E° = 0.7996 V </u>

Ni + 2Ag^+ → Ni^2+ + 2Ag     E° = 1.0566  V

To three significant figures, the standard potential for the cell is 1.06 V .

8 0
4 years ago
A student has the following data recorded: final readings: 760. mm Hg, 6.0L, 197 'C. initial readings:
irina [24]
Is that Geometry or Algebra?
7 0
3 years ago
A state government is trying to decide between using windmills or a nuclear
Gwar [14]

Answer:

B. NUCLEAR POWER CAN PRODUCE ELECTRICITY IN A Y KIND IF WEATHER.

6 0
2 years ago
3. The following data of decomposition reaction of thionyl chloride (SO2Cl2) were collected at a certain temperature and the con
KonstantinChe [14]

Answer:

a) First-order.

b) 0.013 min⁻¹

c) 53.3 min.

d) 0.0142M

Explanation:

Hello,

In this case, on the attached document, we can notice the corresponding plot for each possible order of reaction. Thus, we should remember that in zeroth-order we plot the concentration of the reactant (SO2Cl2 ) versus the time, in first-order the natural logarithm of the concentration of the reactant (SO2Cl2 ) versus the time and in second-order reactions the inverse of the concentration of the reactant (SO2Cl2 ) versus the time.

a) In such a way, we realize the best fit is exhibited by the first-order model which shows a straight line (R=1) which has a slope of -0.0013 and an intercept of -2.3025 (natural logarithm of 0.1 which corresponds to the initial concentration). Therefore, the reaction has a first-order kinetics.

b) Since the slope is -0.0013 (take two random values), the rate constant is 0.013 min⁻¹:

m=\frac{ln(0.0768)-ln(0.0876)}{200min-100min} =-0.0013min^{-1}

c) Half life for first-order kinetics is computed by:

t_{1/2}=\frac{ln(2)}{k}=\frac{ln(2)}{0.013min^{-1}}  =53.3min

d) Here, we compute the concentration via the integrated rate law once 1500 minutes have passed:

C=C_0exp(-kt)=0.1Mexp(-0.013min^{-1}*1500min)\\\\C=0.0142M

Best regards.

6 0
4 years ago
For the reaction A (g) → 2 B (g), K = 14.7 at 298 K. What is the value of Q for this reaction at 298 K when ∆G = -20.5 kJ/mol?
harina [27]

Answer:

Q= 245 =2.5 * 10^2

Explanation:

ΔG = ΔGº + RTLnQ, so also ΔGº= - RTLnK

R= 8,314 J/molK, T=298K

ΔGº= - RTLnK = - 6659.3 J/mol = - 6.7 KJ/mol

ΔG = ΔGº + RTLnQ → -20.5KJ/mol = - 6.7 KJ/mol + 2.5KJ/mol* LnQ

→ 5.5 = LnQ → Q= 245 =2.5 * 10^2

6 0
4 years ago
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