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nlexa [21]
3 years ago
11

Write the half-reactions as they occur at each electrode and the net cell reaction for this electrochemical cell containing tin

and silver: sn(s)|sn2 (aq)||ag (aq)|ag(s)
Chemistry
1 answer:
mojhsa [17]3 years ago
8 0
  The  half  reactions    as  they  occur   at  each  electrode

is  as  follows

at  the  anode  Sn(s)  =sn^2+(aq)  +  2e  -
at  the  cathode  2  ag^+(aq)   + 2e -  =  2Ag (s)

net  cell  reaction  =   Sn (s)  +  2Ag^+(aq)  =  sn^2+ (aq) +  2  Ag  (s)
You might be interested in
The half life of radium-226 is 1600 years. If you have 200 grams of radium today how many grams would be present in 8000 years?
Sveta_85 [38]

Answer:

Half life is the time taken by a radio active isotope to reduce by half of its original amount. Radium-226 has a half life of 1602 years meaning that it would take 1602 years for a mass of radium to reduce by half.

Number of half lives in 9612 years = 9612/1602 = 6 half lives

New mass = Original mass x (1/2)n where n is the number of half lives.

Therefore, New mass= 500 x (1/2)∧6

                                 = 500 x 0.015625

                                 = 7.8125 g

Hence the mass of radium after 9612 years will be 7.8125 grams.

Explanation:

3 0
3 years ago
Read 2 more answers
Calculate the molarity of 29.0g of ethanol (C2H5OH) in 545 mL of solution.
BigorU [14]
Molar mass ethanol:

C2H5OH = 12 x 2+ 1 x 5 + 16  + 1 =  46.0 g/mol

volume = 545 mL in liters: 545 / 1000 => 0.545 L

number of moles:

29.0 / 46.0 => 0.6304 moles

M = n / V

M = 0.6304 / 0.545

M = 1.156 mol/L

hope this helps!


5 0
3 years ago
Read 2 more answers
When a reaction forms between atoms with ground state electron configuration 1s22s2 and 1s22s22p5 the predominant type of bond f
Arisa [49]
C. Ionic is the answer
6 0
2 years ago
Calculate the frequency of the n=2 line in the lyman series of hydrogen
Alona [7]

Answer:

Approximately 2.47\times 10^{15}\; \rm Hz.

Explanation:

The Lyman Series of a hydrogen atom are due to electron transitions from energy levels n \ge 2 to the ground state where n = 1. In this case, the electron responsible for the line started at n = 2 and transitioned to

A hydrogen atom contains only one electron. As a result, Bohr Model provides a good estimate of that electron's energy at different levels.

In Bohr's Model, the equation for an electron at energy level n (

\displaystyle - \frac{k\, Z^2}{n^2} (note the negative sign in front of the fraction,)

where

  • k = 2.179 \times 10^{-18}\; \rm J is a constant.
  • Z is the atomic number of that atom. Z = 1 for hydrogen.
  • n is the energy level of that electron.

The electron that produced the n = 2 line was initially at the

\begin{aligned} &E_{n = 2} \cr &= -\frac{k\, Z^2}{n^2} \cr &= -\frac{2.179 \times 10^{-18} \times 1}{2^2} \cr & \approx -5.4475\times 10^{-19}\; \rm J\end{aligned}.

The electron would then transit to energy level n = 1. Its energy would become:

\begin{aligned} &E_{n = 1} \cr &= -\frac{k\, Z^2}{n^2} \cr &= -\frac{2.179 \times 10^{-18} \times 1}{1^2} \cr & \approx -2.179 \times 10^{-18} \; \rm J\end{aligned}.

The energy change would be equal to

\begin{aligned}&\text{Initial Energy} - \text{Final Energy} \cr &= E_{n = 2} - E_{n = 1} \cr &= -5.4475 \times 10^{-19} - \left(-2.179 \times 10^{-18}\right) \cr & \approx 1.63425\times 10^{-18}\; \rm J \end{aligned}.

That would be the energy of a photon in that n = 2 spectrum line. Planck constant h relates the frequency of a photon to its energy:

E = h \cdot f, where

  • E is the energy of the photon.
  • h \approx 6.62607015\times 10^{-34}\; \rm J \cdot s is the Planck constant.
  • f is the frequency of that photon.

In this case, E \approx 1.63425 \times 10^{-18}\; \rm J. Hence,

\begin{aligned} f &= \frac{E}{h} \cr &\approx \frac{1.63425\times 10^{-18}}{6.62607015\times 10^{-34}} \cr & \approx 2.47 \times 10^{15}\; \rm s^{-1}\end{aligned}.

Note that 1 \; \rm Hz = 1 \; \rm s^{-1}.

6 0
3 years ago
Methanol has a normal boiling point of 64.6C and a heat of vaporization of 35.2 kJ/mol. What is the vapor pressure (in Torr) of
DENIUS [597]

Answer:

vapor pressure of methanol at 12.0C = 75.09 torr

Explanation:

Using Clausius Clapeyron equation

, we have that

ln (P2/P1)= (ΔHvap /R) (1/T1 - 1/T2)

Given

At Normal boiling point,

Temperature T1= 64.6°C = 64.6 + 273 = 337.6 K, Pressure,P1 = 1 atm

Heat of vaporization  = 35.2 kJ/mol

Changing to  J/mol

=35.2 x 1000= 35200 J/mol

Temperature , T2 = 12.0oC = 12 + 273 = 285 K

Using gas constant, R = 8.314 J/mol.K

ln (P2/P1)= -(ΔHvap /R) (1/T1 - 1/T2)

ln (P2/ 1 atm) = (35200 J/mol/ (8.314 J/mol.K) X( 1/337.6 - 1/285)

ln (P2/ 1 atm) =4,233.822 X (0.00296-0.003508)

ln (P2/ 1 atm)  = 4,233.822468  x-0.0005466866

ln (P2/ 1 atm)=  -2.31457

P2 = e^⁻2.31457 x 1 atm

P2=0.098808atm

= 0.098808atm  x760 = 75.09 torr

7 0
3 years ago
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