What is the density of krypton gas (MM = 83.8 g/mol) at 3.00 atm and 100. °C? g/L

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

Which best describes the relationship between the terms frequency, wavelength, and hertz?

Oksanka [162]

Answer:

The correct answer is - Frequency is the number of wavelengths, which is measured in hertz.

Explanation:

Frequency is the number of waves that go through a fixed point at a particular time. Hertz is the SI unit for frequency which means that one hertz is equal to a unit number of waver passes in a unit time to a fixed point.

As the frequency of a wave increases which means the number of waves increases in the unit time, the shorter the wavelength will be.

a higher frequency wave has more energy than a lower frequency wave with the same amplitude.

5 0

3 years ago

Help plsssssss... 3C(s) + 2Fe2O3 (5) ► Fe

blondinia [14]

Answer:

single displacement

Explanation:

the Fe was switched out with the C. they are the only ones that switched, making it a single displacement

6 0

3 years ago

Read 2 more answers

State which of the following set of quantum number would be possible and which would be permissible for an electron in an atom.

raketka [301]

Answer:

The correct option is: e) n=4, l=3, $m_{l}$ = - 2 , $m_{s}$ = - 1/2

Explanation:

The four quantum numbers that describe the electrons in an atom are: The principal: n, azimuthal: l, magnetic: m, and spin: s.

For a given electron, the values of the quantum numbers should be in the given range- principal quantum number: n ≥ 1;

azimuthal quantum number: 0 ≤ l ≤ n − 1;

magnetic quantum number: −l ≤ $m_{l}$ ≤ +l;

spin quantum number: −s ≤ $m_{s}$ ≤ +s

Now, for n= 4,

The value of l: 0 to n − 1 = 0 to 3 = 0, 1, 2, 3

The value of $m_{l}$ for l= 3 : -3 to +3 = -3, -2, -1, 0, +1, +2, +3

The value of $m_{s}$ : -1/2, +1/2

Therefore, the given set of quantum numbers: n=4, l=3, $m_{l}$ = - 2 , $m_{s}$ = - 1/2; are possible and permissible for an electron in an atom.

5 0

3 years ago

A reaction has a rate constant of 1.15 x 10^−2 /s at 400K and 0.685 /s at 450K.

n200080 [17]

Answer:

a) the activation barrier = 122.3 kJ/mol

b) The rate constant at 425 K = 0.1001 /s

Explanation:

Step 1: Data given

Rate constant k1 = 1.15 * 10^−2 /s at 400K (= T1)

Rate constant k2 = 0.685 /s at 450K (=T2)

Step 2: Determine the activation barrier for the reaction.

To determine the activation energy we will use the two-point Arrhenius equation:

ln(k₂/k₁) = (Ea/R)((1/T1) - (1/T2))

⇒ with Ea = the activating energy

⇒ with R = the gas constant = 8.314 J/mol* K

⇒ with k1 = rate constant 1 = 1.15 *10^-2 /s

⇒ with T1 = Temperature 1 = 400 K

⇒ with k2 = rate constant 2 = 0.685/s

⇒ with T2 = temperature 2 = 450 K

= - (Ea/R)(T₁ - T₂)/T₁T₂

Ea = (R*ln (k2/k1)) / ((1/T1)- (1/T2))

Ea = (8.314* ln(0.685/0.0115)) / ((1/400) - (1/450))

Ea = 122327.6 = 122.3 kJ/mol

B) What is the value of the rate constant at 425 K

For rate constant at 425 K.

Substitute the value of activation energy as 122327.6 J/mol, initial temperature as 400 K, final temperature as 425 K, rate constant at 400 K

1/T1 - 1/ T3 = 1/400 - 1 /425 = 1.47*10^-4

⇒ with T1 = the initial temperature = 400 K

⇒ with k1 = the rate constant at 400 K = 1.15 * 10^-2 /s

⇒ with T3 = the nex temperature = 425 K

⇒ with k3 = the rate constant at 425 K

ln(k3/k1) = Ea/R * ((1/T1)- (1/ T3))

⇒ with k3 = the rate constant at 425 K

⇒ with T3 = 425 K

k3/k1 = e^(Ea/R * ((1/T1)- (1/ T3)))

k3 = k1* e^(Ea/R * ((1/T1)- (1/ T3)))

k3 = 0.0115 * e^(122327.6/8.314 * (1.4710^-4))

k3 = 0.0115* e^2.1643

k3 = 0.1001 /s

4 0

3 years ago