Constant = 8.314JK⁻¹mol⁻¹
T is for temperature which is 298K
Faraday constant value is 96500C/mol
n is the number of electrons which are transferred in the reaction.
Ecell = E₀cell - RT/nFiN [cathode]/[anode]
Ecell = E₀cell - RT/nF In [PH₂]/[H⁺]²
Ecell = 0.00-8.314 JK⁻¹ mol⁻¹ × 298k/ 2× 96500C/mol In [2.4atm]/ 0.70]²
Ecell = 0.00 - 0.0129 In (2.59)
Ecell = 0.00 - 0.0129 × 0.951
Ecell = -0.0122V
∴Ecell is = -o.0122v

7 0

3 years ago

A child on a sled is perched at the top of a hill, at a height 7.0 m above the base of the hill. The total mass of the girl plus

spin [16.1K]

Answer:

a)11,71 m/s

b)12.73 m/s

Explanation:

We apply the conservation principle because there is no friction between the hill and the sled:

Total initial energy (Ei) = Total initial energy (Ef), Equation(1)

We define:

K = Kinetic energy = $\frac{1}{2}*m*v^{2}$

U = Potential energy = $m*g*h$

m = mass (kg)

v = velocity (m/s)

h = height (m)

g = gravity acceleration = $9.8m/s^{2}$

A) hi=7m, hf=0 ,vi=0 , vf=?

$Ei=Ef\\Ki+Ui=Kf+Uf$

$1/2*m*vi^{2} +m*g*hi= 1/2*m*vf^{2} +m*g*hf$

$0+m*g*hi=1/2*m*vf^{2}+0$

$(2*m*g*hi)/m=vf^{2}$

$vf=\sqrt{2*g*hi}$

$vf=\sqrt{2*9.8*7}$

$vf=11.71 m/s$

Answer: She reaches the base of the hill with speed of 11.22 m/s

B) hi=7 m, vi=5 m/s, hf=0, vf=?

$m*g*hi+1/2*m*vi^{2} =m*g*hf+1/2*m*vf^{2}$

We divide all terms by m

$g*hi+1/2vi^{2} =g*hf+1/2*vf^{2}$

$9.8*7+1/2*5^{2} =1/2*vf^{2}$

$2(68.6+12.5)=vf^{2}$

$162.2=vf^{2}$

$vf=\sqrt{162.2}$

$vf=12.73 m/s$

Answer: She will move with a speed of 12.73 m/s when she reaches the base of the hill

3 0

3 years ago

How much work is done by a crane that lowers 1000N of material a distance of 150?

Norma-Jean [14]

The formula is ''w=fd''

So according this fromula we excersise:

W= (1000N) (150)

So we multiply ''1000'' times ''150''

W= 150,000J

Hope i helped! If you need anything else ask me! :)

7 0

3 years ago

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