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3 years ago

9. During vaporization, a substance changes from a liquid to a(n) _______________. (1 point)

Chemistry

1 answer:

Crank3 years ago

3 0

Gas. Is the answer to the question.

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Calculate the silver ion concentration in a saturated solution of silver(i) sulfate (ksp = 1.4 × 10–5).

deff fn [24]

Answer:

= 0.030 M

Explanation:

We can take x to be the concentration in mol/L of Ag2SO4 that dissolves

Therefore; concentration of Ag+ is 2x mol/L and that of SO4^2- x mol/L.

Ksp = 1.4 x 10^-5

Ksp = [Ag+]^2 [SO42-]

= (2x)^2(x)

= 4x^3

Thus;

4x^3 = 1.4 x 10^-5

= 0.015 M

molar solubility = 0.015 M

But;

[Ag+]= 2x

Hence; silver ion concentration is

= 2 x 0.015 M

= 0.030 M

6 0

2 years ago

Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e

notsponge [240]

Answer: The concentration of $Sn^{2+}$ in the cell is $9.0\times 10^{-3}M$

Explanation:

We are given:

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-$ $E^o_{Zn^{2+}/Zn}=-0.76V$

Reduction half reaction: $Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)$ $E^o_{Sn^{2+}/Sn}=-0.136V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=-0.136-(-0.76)=0.624V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = 0.660 V

$E^o_{cell}$ = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=2.5\times 10^{-3}M$

$[Sn^{2+}]$ = ?

Putting values in above equation, we get:

$0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})$

$[Sn^{2+}]=9.0\times 10^{-3}M$

Hence, the concentration of $Sn^{2+}$ ions is $9.0\times 10^{-3}M$

3 0

2 years ago

Which statement best describes the polarity of the molecule Sil4?

Nadusha1986 [10]

Answer:

I dont a picture

Explanation:

but the answer is 24

6 0

2 years ago

Which pair of dispersed phases and dispersing media can never form a colloid?

lana66690 [7]

Answer:

Option B Liquid and Gas

7 0

2 years ago

To find the atomic mass of a particular atom you need to know

AveGali [126]

The answer is (C) as the definitions states that atomic mass is the number of protons and neutrons.
Hope this helps :).

6 0

2 years ago