The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is suddenly applied to the initially uncharged plates through a 975 ohm resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 75 volts

23.92 ns

Explanation:

The capacitance, C = (e(0) * A) / d

C = [8.85*10^-12 * 2*10^-2 * 10*10^-2] / 1*10^-3

C = (1.77*10^-14) / 1*10^-3

C = 1.77*10^-11

C = 17.7 pF

Vc = V * [1 - e^-(t/CR)]

75 = 100 * [1 - e^-(t/CR)]

75/100 = 1 - e^-(t/CR)

e^-(t/CR) = 1 - 0.75

e^-(t/CR) = 0.25

If we take the log of both sides, we then have

Log e^-(t/CR) = Log 0.25

-t/CR = In 0.25

t = -CR In 0.25

t = - 17.7*10^-12 * 975 * -(1.386)

t = 2.392*10^-8 s

t = 23.92 ns

  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