9. Two 6 μC point charges are located at the points (2 m, 0 m) and (0 m, 2 m), and a -3.5 μC charge is at the origin. How much work is required to move these charges an infinite distance away from each other?

-27 mJ

Explanation:

The work done in moving the three charges an infinite distance from each other is the electric potential energy of the system. Let q₁ = q₂ = 6 μC = 6 × 10⁻⁶ C, q₃ =  -3.5 μC = -3.5 × 10⁻⁶ C. Let r₁ be the distance between q₁ and q₃. r₁ = 2 m. Let r₂ be the distance between q₂ and q₃. r₂ = 2 m. Let r₃ be the distance between q₁ and q₂. r₃ = √(2 - 0)² + (0 - 2)² = √4 = 2 m

The potential energy is thus U

U = kq₁q₂/r₃ + kq₁q₃/r₁ + kq₂q₃/r₂

Since q₁ = q₂ = q and r₁ = r₂ = r₃ = r

U = kq(q + 2q₃)/r =

U = 9 × 10⁹ Nm²/C² × 6 × 10⁻⁶ C(6 × 10⁻⁶ C + 2 × -3.5 × 10⁻⁶ C)/2 m

U = 9 × 10⁹ Nm²/C² × 6 × 10⁻⁶ C(6 × 10⁻⁶ C - 7 × 10⁻⁶ C)/2 m

U = 9 × 10⁹ Nm²/C² × 6 × 10⁻⁶ C(- 1 × 10⁻⁶ C)/2 m

U = -54 × 10⁻³ Nm²/2 m

U = -27 × 10⁻³ Nm

U = -27 × 10⁻³ J

U = -27 mJ

So a work of -27 mJ must be done to move the charges an infinite distance away from each other

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂

0.55x745.7j and a calculator