What is the percent copmosition of chromium in CrF3

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

3 years ago

Calculate the vapor pressure of water at T=90°C

LenKa [72]

The equilibrium vapour pressure is typically the pressure exerted by a liquid .... it is A FUNCTION of temperature...

Explanation:

By way of example, chemists and physicists habitually use

P

saturated vapour pressure

...where

P

SVP

is the vapour pressure exerted by liquid water. At

100

∘

C

,

P

SVP

=

1

⋅

a

t

m

. Why?

Well, because this is the normal boiling point of water: i.e. the conditions of pressure (i.e. here

1

⋅

a

t

m

) and temperature, here

100

∘

C

, at which the VAPOUR PRESSURE of the liquid is ONE ATMOSPHERE...and bubbles of vapour form directly in the liquid. As an undergraduate you should commit this definition, or your text definition, to memory...

At lower temperatures, water exerts a much lower vapour pressure...but these should often be used in calculations...especially when a gas is collected by water displacement. Tables of

saturated vapour pressure

are available.

4 0

3 years ago

How many grams of mercury would be contained in 15 compact fluorescent light bulbs?

zhenek [66]

Answer:

Grams of mercury= 0.06 g of Hg

Note: The question is incomplete. The complete question is as follows:

A compact fluorescent light bulb contains 4 mg of mercury. How many grams of mercury would be contained in 15 compact fluorescent light bulbs?

Explanation:

Since one fluorescent light bulb contains 4 mg of mercury,

15 such bulbs will contain 15 * 4 mg of mercury = 60 mg

1 mg = 0.001 g

Therefore, 60 mg = 0.001 g * 60 = 0.06 g of mercury.

Compact fluorescent lightbulbs (CFLs) are tubes containing mercury and noble gases. When electricity is passed through the bulb, electron-streams flow from a tungsten-coated coil. They collide with mercury atoms, exciting their electrons and creating flashes of ultraviolet light. A phosphor coating on the inside of the tube absorbs this UV light flashes and re-emits it as visible light. The amount of mercury in a fluorescent lamp varies from 3 to 46 mg, depending on lamp size and age.

5 0

3 years ago

Answer question 13. What is the mechanical advantage of a lever that can lift a 100 N load with a input force of 20 N

vazorg [7]

MA= output force/ input force
MA= 100N/20N
MA= 50

8 0

3 years ago

Is flammability a form of chemical change or physical change

Vinvika [58]

Chemical change C===3

6 0

3 years ago

Read 2 more answers