What is the amount of electrical energy utilized in moving the electrons from one terminal point in

3 answers
Question:

what is the amount of electrical energy utilized in moving the electrons from one terminal point in a battery called

Answers

Explanation:

In a similar way charged particles such as electrons need to have work done to move them from one point to another. The amount of work per unit of charge is called is called the electric potential difference between the two points. The unit of potential difference is called the volt.

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence,   the electronegativity (or the ability to attract electrons) is lessened."

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

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