5. A charge moves a distance of 2.0 cm in the direction of a uniform electric field having a magnitude of 215 N/C. The electrical potential energy of the charge decreases by 6.9 x10-19 J as it moves. Find the magnitude of the charge on the moving particle. (Hint: The electrical potential energy depends on the distance moved in the direction of the field.)

The magnitude of charge on the particle is .        

Explanation:

Given that,

The magnitude of electric field, E = 215 N/C

The electrical potential energy of the charge decreases by, as it moves.

We need to find the magnitude of the charge on the moving particle. The change in electric potential energy is given by :

.........(1)

The electric potential in terms of electric field is given by :

Equation (1) becomes :

So, the magnitude of charge on the particle is .                                    

0.863333

86%

hope this

  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