The combustion of hydrogen and oxygen is commonly used to

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

If 0.0106 g of a gas dissolves in 0.792 L of water at 0.321 atm, what quantity of this gas (in grams) will dissolve at 5.73 atm?

Alex73 [517]

Answer:

0.189 g.

Explanation:

This problem is an application on Henry's law.

Henry's law states that the solubility of a gas in a liquid is directly proportional to its partial pressure of the gas above the liquid.

Solubility of the gas ∝ partial pressure

If we have different solubility at different pressures, we can express Henry's law as:

S₁/P₁ = S₂/P₂,

S₁ = 0.0106/0.792 = 0.0134 g/L and P₁ = 0.321 atm

S₂ = ??? g/L and P₂ = 5.73 atm

So, The solubility of the gas at 5.73 atm (S₂) = S₁.P₂/P₁ = (0.0134 g/L x 5.73 atm) / (0.321 atm) = 0.239 g/L.

The quantity in (g) = S₂ x V = (0.239 g/L)(0.792 L) = 0.189 g.

8 0

2 years ago

A topographical map would be useful for which activity? A. planning a hiking trip B. studying plant growth over time C. studying

amm1812

The answer is A. planning a hiking trip

A topographical map would not help studying plant growth over time unless you are looking for a better altitude to plant said plants.

A topographical map would not help studying rainfall for one year unless your location was really so high or low that it affected your weather

And most of all, a topographical map would not be useful for planning a cuise across the Atlantic Ocean because the elevation of the sea is zero!

A topographical map would be useful for planning a hiking trip because there are many factors and details that a hike should have. Which includes height, distance, paths, and elevation.

5 0

3 years ago

The symbol for the element in period 2, group 13

denis-greek [22]

Element: Boron
Symbol: B
So answer is B

8 0

2 years ago

when an electron moves from a higher orbit to a lower one does it always follow the same path explained

Ilia_Sergeevich [38]

Yes, electron follows the same path when it absorb and loses energy.

Yes, when an electron moves from a higher orbit to a lower orbit it always follow the same path as it moves from a lower orbit to a higher orbit. When electron absorb energy it has the power to move from lower orbit to higher orbit or energy level.

While on the other hand, when an electron loses that energy, it comes back to its original position from which it moves earlier when it absorb energy so we conclude that electron follows the same path when it absorb and loses energy.

Learn more: brainly.com/question/24962163

8 0

2 years ago

Read 2 more answers