An alpha particle (charge 2e) is sent at high speed toward an atomic nucleus. The electric force acting on the alpha particle is 79.8 N when it is 2.29 x 10(-14) m away from the nucleus. What is the charge on the atomic nucleus? (e = 1.60 x 10(-19) C, Kc = 9 x 10(9) N•m(2)/c(2)

1.45×10⁻¹⁶ C

Explanation:

Applying,

F = kqq'/r² Equation 1

Where F = force acting on the alpha particle, k = coulomb's constant, q = charge on the alpha particle, q' = charge on the atomic nucleus, r = distance of seperation between the alpha particle and the atomic nucleus.

make q' the subject of the equation

q' = Fr²/(kq) Equation 2

From the question,

Given: F = 79.8 N, r = 2.29×10⁻¹⁴ m, q = 2e = (2×1.6×10⁻¹⁹) = 3.2×10⁻¹⁹ C, k = 9×10⁹ Nm²/C²

Substitute these values into equation 2

q' = [79.8×(2.29×10⁻¹⁴)²]/(3.2×10⁻¹⁹×9×10⁹)

q' = (4.18×10⁻²⁶)/(2.88×10⁻¹⁰)

q' = 1.45×10⁻¹⁶ C

  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