Consider the electrostatic force between two particles of equal mass m = 32 × 10^-26 kg and carrying different charges q1 = 17e and q2 = 28e, where e denotes the magnitude of the charge on an electron. at what distance, in meters, is the electrostatic force between these two particles equal to the weight of one of them?

0.187 m

Explanation: In order to explain this problem we have to consider the Coulomb and gravitational force, which are given by;

Electric force

Fe=(k*q1*q2)/d^2 where k is a constant equal to 9*10^9 N/C^2*m^2 and d is the distanvce between the charges.

gravitational force

Fg=me*g  

equaling both forces,  we have:

(k*q1*q2)/d^2 = me*g  

d^2= (k*q1*q2)/me*g = 9*10^9*17*28*(1.6*10^-19)^2/(32*10^-26*9.8)=

=0.187 m

  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

are you like autistic or something?

xd