Ask question

Ask question

All categories

kondor19780726 [428]

3 years ago

10

328. mL of 0.00345 M NaI (aq) is combined with 703. mL of 0.00802 M Pb(NO3)2 (aq). Determine if a precipitate will form given th

at the Ksp of Pbl2 is 1.40x10-8.

1 answer:

Vsevolod [243]3 years ago

6 0

Answer:

The solution will not form a precipitate.

Explanation:

The Ksp of PbI₂ is:

PbI₂(s) ⇄ 2I⁻(aq) + Pb²⁺(aq)

Ksp = 1.40x10⁻⁸ = [I⁻]²[Pb²⁺] <em>Concentrations in equilibrium</em>

When 328mL of 0.00345M NaI(aq) is combined with 703mL of 0.00802M Pb(NO₃)₂. Molar concentration of I⁻ and Pb²⁺ are:

[I⁻] = 0.00345M × (328mL / (328mL+703mL) =<em> 1.098x10⁻³M</em>

[Pb²⁺] = 0.00802M × (703mL / (328mL+703mL) =<em> 5.469x10⁻³M</em>

<em />

Q = [I⁻]²[Pb²⁺] <em>Concentrations not necessary in equilibrium</em>

If Q = Ksp, the solution is saturated, Q > Ksp, the solution will form a precipitate, if Q < Ksp, the solution is not saturated.

Replacing:

Q = [1.098x10⁻³M]²[5.469x10⁻³M] = 6.59x10⁻⁹

As Q < Ksp, the solution is not saturated and <em>will not form a precipitate</em>.

You might be interested in

When lewis structures are drawn, do you only show valence electrons??

Lynna [10]

Yes you do cause its more explanation to
<span />

8 0

3 years ago

What products result from mixing aqueous solutions of Cr(NO3)2(aq) and NaOH(aq)? Question 10 options: Cr(OH)2(s), Na+(aq), and N

Alecsey [184]

Answer:

Cr(OH)2(s), Na+(aq), and NO3−(aq)

Explanation:

Let is consider the molecular equation;

2NaOH(aq) + Cr(NO3)2(aq) -----> 2NaNO3(aq) + Cr(OH)2(s)

This is a double displacement or double replacement reaction. The reacting species exchange their partners. We can see here that both the sodium ion and chromium II ion both exchanged partners and picked up each others partners in the product.

Sodium ions and nitrate ions now remain in the solution while chromium II hydroxide which is insoluble is precipitated out of the solution as a solid hence the answer.

4 0

3 years ago

Aclosed system contains an equimolar mixture of n-pentane and isopentane. Suppose the system is initially all liquid at 120°C an

garri49 [273]

Explanation:

The given data is as follows.

T = $120^{o}C$ = (120 + 273.15)K = 393.15 K,

As it is given that it is an equimolar mixture of n-pentane and isopentane.

So, $x_{1}$ = 0.5 and $x_{2}$ = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

$p^{sat}_{1}$ (393.15 K) = 9.2 bar

$p^{sat}_{1}$ (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

$p_{1} = x_{1} \times p^{sat}_{1}$

= $0.5 \times 9.2 bar$

= 4.6 bar

and, $p_{2} = x_{2} \times p^{sat}_{2}$

= $0.5 \times 10.5 bar$

= 5.25 bar

Therefore, the bubble pressure will be as follows.

P = $p_{1} + p_{2}$

= 4.6 bar + 5.25 bar

= 9.85 bar

Now, we will calculate the vapor composition as follows.

$y_{1} = \frac{p_{1}}{p}$

= $\frac{4.6}{9.85}$

= 0.467

and, $y_{2} = \frac{p_{2}}{p}$

= $\frac{5.25}{9.85}$

= 0.527

Calculate the dew point as follows.

$y_{1}$ = 0.5, $y_{2}$ = 0.5

$\frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}$

$\frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}$

$\frac{1}{P}$ = 0.101966 $bar^{-1}$

P = 9.807

Composition of the liquid phase is $x_{i}$ and its formula is as follows.

$x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{9.2}$

= 0.5329

$x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{10.5}$

= 0.467

4 0

3 years ago

How many electrons in an atom can share the quantum numbers n = 4 and l = 3?

o-na [289]

Answer:

$\boxed{\text{14}}$

Explanation:

If l = 3, the electrons are in an f subshell.

The number of orbitals with a quantum number l is 2l + 1, so there

are 2×3 + 1 = 7 f orbitals.

Each orbital can hold two electrons, so the f subshell can hold 14 electrons.

$\boxed{\textbf{14 electrons}} \text{ can share the quantum numbers n = 4 and l = 3.}$

3 0

4 years ago

Tengo que resolver con procedimiento

zzz [600]

Answer:

Distance = 200 km

Distance = 204 km

Speed = 77 km/h

Time = 21.42 h

Explanation:

Given:

A.

Speed = 100 km/h , Time = 2 h

Find:

Distance

B.

Speed = 68 km/h , Time = 3 h

Find:

Distance

C.

Distance = 154 km , Time = 2 h

Find:

Speed

D.

Distance = 1500 km speed = 70 km/h

Find:

Time

Computation:

Speed = distance / time

A.

Distance = 100 x 2

Distance = 200 km

B.

Distance = 68 x 3

Distance = 204 km

C.

Speed = 154 / 2

Speed = 77 km/h

D.

Time = 1500 / 70

Time = 21.42 h

6 0

3 years ago