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baherus [9]
3 years ago
13

Write the cell notation for an electrochemical cell consisting of an anode where Mn (s) is oxidized to Mn2 (aq) and a cathode wh

ere Co2 (aq) is reduced to Co (s) . Assume all aqueous solutions have a concentration of 1 mol/L.
Chemistry
1 answer:
morpeh [17]3 years ago
8 0

Answer:

Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)

Explanation:

In writing the cell notation for an electrochemical cell, the anode is written on the left hand side while the cathode is written on the right hand side. The two half cells are separated by two thick lines which represents the salt bridge.

For the cell discussed in the question; the Mn(s)/Mn^2+(aq) is the anode while the Co^2+(aq)/Co(s) half cell is the cathode.

Hence I can write; Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)

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What is the volume of 0.4 mole of gas at stp? 30 points!
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Answer:

89.6 liters

Explanation:

A STP (standard temperature and pressure) ONE mole of any idea gas will occupies 22.4 liters. So,...

4 moles x 22.4 L/mol = 89.6 liters

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A certain radioactive isotope decays at a rate of 0.2​% annually. Determine the ​half-life of this​ isotope, to the nearest year
pychu [463]

Answer:

The half-life of the radioactive isotope is 346 years.

Explanation:

The decay rate of the isotope is modelled after the following first-order linear ordinary differential equation:

\frac{dm}{dt} = -\frac{m}{\tau}

Where:

m - Current isotope mass, measured in kilograms.

t - Time, measured in years.

\tau - Time constant, measured in years.

The solution of this differential equation is:

m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }

Where m_{o} is the initial mass of the isotope. It is known that radioactive isotope decays at a yearly rate of 0.2 % annually, then, the following relationship is obtained:

\%e = \frac{m(t)-m(t+1)}{m(t)}\times 100\,\% = 0.2\,\%

1 - \frac{m(t+1)}{m(t)} = 0.002

1 - \frac{m_{o}\cdot e^{-\frac{t+1}{\tau} }}{m_{o}\cdot e^{-\frac{t}{\tau} }}=0.002

1 - e^{-\frac{1}{\tau} } = 0.002

e^{-\frac{1}{\tau} } = 0.998

-\frac{1}{\tau} = \ln 0.998

The time constant associated to the decay is:

\tau = -\frac{1}{\ln 0.998}

\tau \approx 499.500\,years

Finally, the half-life of the isotope as a function of time constant is given by the expression described below:

t_{1/2} = \tau \cdot \ln 2

If \tau \approx 499.500\,years, the half-life of the isotope is:

t_{1/2} = (499.500\,years)\cdot \ln 2

t_{1/2}\approx 346.227\,years

The half-life of the radioactive isotope is 346 years.

6 0
3 years ago
On the galapagos islands, Darwin studied many creatures, including finches. Explain what he observed and how it relates to natur
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Hope this helped! Have a great day! 
5 0
3 years ago
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