A balloon is inflated with 1.25 L of helium at a temperature of 298K. What is

The standard reduction potentials for the Ag+|Ag(s) and Zn2+| Zn(s) half-cell reactions are +0.799 V and -0.762 V, respectively.

mihalych1998 [28]

Answer: The potential of the given cell is 1.551 V

Explanation:

The given chemical cell follows:

$Zn(s)|Zn^{2+}(0.125M)||Ag^{+}(0.240M)|Ag(s)$

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}(0.125M)+2e^-;E^o_{Zn^{2+}/Zn}=-0.762V$

Reduction half reaction: $Ag^{+}(0.240M)+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.799V$ ( × 2)

Net cell reaction: $Zn(s)+2Ag^{+}(0.240M)\rightarrow Zn^{2+}(0.125M)+2Ag(s)$

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:

$E^o_{cell}=0.799-(-0.762)=1.561V$

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

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

where,

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

$E^o_{cell}$ = standard electrode potential of the cell = +1.561 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=0.125M$

$[Ag^{+}]=0.240M$

Putting values in above equation, we get:

$E_{cell}=1.561-\frac{0.059}{2}\times \log(\frac{(0.125)}{(0.240)^2})$

$E_{cell}=1.551V$

Hence, the potential of the given cell is 1.551 V

6 0

4 years ago

As your rise upwards in the atmosphere, air pressure.

ozzi

Increases. It gets harder to breathe and keep the rhythm.

3 0

3 years ago

How many atoms are there in 20.00 moles of calcium (Ca)?

laila [671]

Answer:

20 x (6.022 x 10^23) atoms in 20 moles of Calcium.

Explanation:

A mole contains 6.022 x 10^23 atoms, which is also known as Avogadro's number, and this number is same for each element. So, there are 20 x (6.022 x 10^23) atoms in 20 moles of Calcium.

7 0

3 years ago

Read 2 more answers

The yearly amounts of carbon emissions from cars in Belgium are normally distributed with a mean of 13.9 gigagrams per year and

zavuch27 [327]

Answer:

The correct option is;

c. 0.167

Explanation:

The parameters given are;

The mean, μ = 13.9 Gigagrams/year

The standard deviation, σ = 5.8 Gigagrams/year

The z-score formula is given as follows;

$z = \dfrac{x - \mu }{\sigma }$

Where:

x = Observed score = 11.5 Gigagrams/year

We have;

$z = \dfrac{11.5 - 13.9 }{5.8 } = \dfrac{-2.4 }{5.8 } = 0.4138$

From the z-score table relations/computation, the probability (p-value) = 0.6605

Where:

x = Observed score = 14 Gigagrams/year

We have;

$z = \dfrac{14- 13.9 }{5.8 } = \dfrac{0.1}{5.8 } = 0.01724$

From the z-score table relations/computation, the probability (p-value) = 0.4931

Therefore, the probability, $p_{ca}$ , that the amount of carbon emissions from cars in Belgium for a randomly selected year are between 15.5 Gigagrams/year and 14 Gigagrams/year = The area under the normal curve bounded by the p-values for the two amounts of carbon emission

Which gives;

$p_{ca}$ = 0.6605 - 0.4931 = 0.1674 ≈ 0.167

Therefore, the correct option is c. 0.167.

4 0

4 years ago

What is the molar mass of a substance?

Blizzard [7]

The molar mass of a substance is 47.88 g/mol because in Titanium there is 47.88 g/mol there is one mole, or 6.022 x 1023titanium atoms.

7 0

4 years ago