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CaHeK987 [17]
3 years ago
8

When the difference in redox potential between two molecules is highly positive, what is true of the transfer of electrons betwe

en them?
Physics
1 answer:
Bess [88]3 years ago
3 0

Answer:

Explanation:

Redox potential is the ability of a chemical species to loose or gain electrons.

In a redox reaction there is the REDUCING agent which undergoes oxidation loosing electrons and the OXIDIZING agents undergoes reduction and gain electrons

What is true of the transferof electrons between them is that because the difference in potential is highly positive be cause of reduction which leads to the gain of electrons

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Use Kepler’s laws and the period of the Moon (27.4 d) to determine the period of an artificial satellite orbiting very near the
miv72 [106K]

As according to Kepler 's law

T =(4π²r³/ Gm)^1/2

here r= distance  from earth center to satellite = 6400km = 6400000m

G = earth's gravitational constant= 6.67×10^-11

m = mass of earth= 5.98 ×10^24

so T =[ { 4× (3.1416)²×(6400000)³}/ {(6.67×10^-11)×( 5.98 ×10^24)} ]^1/2

T= 5133 sec


7 0
3 years ago
Read 2 more answers
In the chemical equation above, the small number after the O in 1202 represent —
Murrr4er [49]

Answer:

G.) The number atoms of that element in the molecule

Explanation:

F is incorrect because the coefficient represents the amount of one type of molecule, not the subscript

G is correct because subscripts represent how many atoms of that element are present in that single molecule

H is incorrect because energy is not represented in this simple type of equation

J is incorrect because it doesn't even make sense

7 0
3 years ago
A carbon rod with a radius of 1.9 mm is used to make a resistor. What length of the carbon rod should be used to make a 3.7 Ω re
vladimir2022 [97]

Answer:

Length = 2.32 m

Explanation:

Let the length required be 'L'.

Given:

Resistance of the resistor (R) = 3.7 Ω

Radius of the rod (r) = 1.9 mm = 0.0019 m [1 mm = 0.001 m]

Resistivity of the material of rod (ρ) = 1.8\times 10^{-5}\ \Omega\cdot m

First, let us find the area of the circular rod.

Area is given as:

A=\pi r^2=3.14\times (0.0019)^2=1.13\times 10^{-5}\ m^2

Now, the resistance of the material is given by the formula:

R=\rho( \frac{L}{A})

Express this in terms of 'L'. This gives,

\rho\times L=R\times A\\\\L=\frac{R\times A}{\rho}

Now, plug in the given values and solve for length 'L'. This gives,

L=\frac{3.7\ \Omega\times 1.13\times 10^{-5}\ m^2}{1.8\times 10^{-5}\ \Omega\cdot m}\\\\L=\frac{4.181}{1.8}=2.32\ m

Therefore, the length of the material required to make a resistor of 3.7 Ω is 2.32 m.

5 0
3 years ago
Why do some scientists think that Irr II galaxies have irregular, distorted shapes?
laila [671]
They believe the distortions happened when two galaxies collided.

Hope This Helps :)
3 0
3 years ago
It is known that the kinetics of recrystallization for some alloy obeys the Avrami equation, andthat the value of n in the expon
trapecia [35]

Answer:8.76\times 10^{-3} min^{-1}

Explanation:

Given

n=5

0.3 fraction recrystallize after 100 min

According to Avrami equation

y=1-e^{-kt^n}

where y=fraction Transformed

k=constant

t=time

0.3=1-e^{-k(100)^5}

e^{-k(100)^5} =0.7

Taking log both sides

-k\cdot (10^{10}=\ln 0.7

k=3.566\times 10^{-11}

At this Point we want to compute t_{0.5}\ i.e.\ y=0.5

0.5=1-e^{-kt^n}

0.5=e^{-kt^n}

0.5=e^{-3.566\times 10^{-11}\cdot (t)^5}

taking log both sides

\ln 0.5=-3.566\times 10^{-11}\cdot (t)^5

t^5=1.943\times 10^{10}

t=114.2 min

Rate of Re crystallization at this temperature

t^{-1}=8.76\times 10^{-3} min^{-1}

3 0
3 years ago
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