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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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Do you have a diagram or anything?

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4 years ago
A bicycle pump contains 200 cm3 of air and is connected to a bicycle tyre. The volume of the tyre is 800 cm3. The pressure of th
Aleksandr [31]

Answer:

The total volume of the air is 1000 cubic centimeters.

Explanation:

Since the bicycle pump and the bicycle tyre have the same pressure, then the total volume of the air is the sum of the volume of each element, then we translate this into the following artihmetical expression:

V = 200\,cm^{3}+800\,cm^{3}

V = 1000\,cm^{3}

The total volume of the air is 1000 cubic centimeters.

7 0
3 years ago
Two charges are located in the xx – yy plane. If ????1=−4.25 nCq1=−4.25 nC and is located at (x=0.00 m,y=1.080 m)(x=0.00 m,y=1.0
Sati [7]

Answer:

Ex=  -17.1 N/C

Ey =  +26.9 N/C

Explanation:

We apply formula of electric field:

Ep=k*q/d²

Ep:  Electric field at point ( N/C)

q: Electric charge (C)

k: coulomb constant (N.m²/C²)

d: distance from charge q to point P (m)

In the attached graph we observe the directions of the electric field at P(0,0) due to q1 and q2

Calculation of the field at point P due to the load q₁

E₁=k*q₁/d₁² = 9*10⁹*4.25*10⁻⁹/1.080²= 32.8 N/C : Magnitude of E1

Direction of E₁ :Because the charge q₁ is negative the field enters the charge (+ y)

Calculation of the field at point P due to the load q₂

d_{2} = \sqrt{1.30^{2}+0.450^{2}  }

d₂=1.375 m

E₂=k*q₂/d₂² = 9*10⁹*3.80*10⁻⁹/ 1.375² = 18.09 N/C Magnitude of E₂

Direction of E₂ :Because the charge q₂ is positive the field leaves the charge in direction of angle β

, then,E₂ tiene componentes x-y  en P.

E₂x=-E₂cos β= -18.09*(1.3/1.375)= -17.1 N/C

E₂y=-E₂sin β= -18.09*(0.45/1.375)= -5.9 N/C

Calculation of the electric field at point P located at the origin(0,0)

Ex=E₂x= -17.1 N/C

Ey=E₁y+E₂y =32.8 N/C -5.9 N/C = 26.9 N/C

4 0
3 years ago
The deflection of alpha particles in rutherford’s gold foil experiments resulted in what change to the atomic model? the additio
skelet666 [1.2K]
The answer is: The addition of a small dense nucleus at the center of the atom.
5 0
3 years ago
Read 2 more answers
The diagram shows two forces of equal magnitude acting on an object. If the common magnitude of the forces is 3.6 N and the angl
Nuetrik [128]
<h3>Answer</h3>

6.6 N pointing to the right

<h3>Explanation</h3>

Given that,

two forces acting of magnitude 3.6N

angle between them = 48°

To find,

the third force that will cause the object to be in equilibrium

<h3>1)</h3>

Find the vertical and horizontal components of the two forces

vertical force1 = sin(24)(3.6)

vertical force2= -sin(24)(3.6)

<em>(negative sign since it is acting on opposite direction)</em>

vertical force3 = sin(24)(3.6) - sin(24)(3.6)

                        = 0

<h3>2)</h3>

horizontal force1 = cos(24)(3.6)

horizontal force2= cos(24)(3.6)

horizontal force3 = cos(24)(3.6) + cos(24)(3.6)

                            = 2(cos(24)(3.6))

                            = 6.5775 N

                            ≈ 6.6 N

<em />

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4 0
4 years ago
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