An AC voltage source, with a peak output of 72 V, results in dissipation of energy in a resistor at rate of 140 W. What is the value of the resistance?

18.51 Ω

Explanation:

Power dissipated in electric circuits is given by P = IV

But, V = IR And I = V/R

P = (V/R) V = V²/R

But this is an AC circuit, and the voltage in the power dissipated equation is the DC voltage.

DC voltage of the AC circuit = Vrms = (Vpeak)/√2 = 72/√2 = 50.912 V

P = V²/R

R = V²/P = 50.912²/140 = 18.51 Ω

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explanation:

  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