Acopper wire of cross sectional area 2.0mm2 carries a current of 10a. how many electrons pass through a given cross section of wire in one second?
a 1.0×10^1. b 5.0×10^6. c 6.3×10^19. d 3.1×10^25

The amount of electron that pass through the wire is 6.3*10^19. (Option C)

Explanation:

The current represent the amount of electriccharge iper period of time. When the current is in Ampere (A) the charge is expressed in Coulombs (C) and the time in seconds (s).

I=Q/T

Solving for Q and replacing the value for the current an a time of 1 seconds:

Q=I.T = 10 A * 1s = 10 C/s * 1s = 10 C

The amount of electrons is calculated by dividing the total charge by the charge of one electron. Recall for the charge of a one electron q is:

q=1.6*10^-19 C

Number of electrons = 10 C  / 1.6*10^-19 C = 6.3*10^19  

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz

the independent variable, or the one that is intentionally manipulated is the speed.