You wish to produce an emf of 41.0 mV using an inductor whose inductance is 13.0 H. You start with a current of 1.50 mA through the inductor and increase the current at a steady rate. What is the current through the inductor at the end of 2.60 s

The current through the inductor at the end of 2.60s is 9.7 mA.

Explanation:

Given;

emf of the inductor, V = 41.0 mV

inductance of the inductor, L = 13 H

initial current in the inductor, I₀ = 1.5 mA

change in time, Δt = 2.6 s

The emf of the inductor is given by;

Therefore, the current through the inductor at the end of 2.60 s is 9.7 mA.

  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

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