A light bulb and a solenoid are connected in series to a battery. An irod rod is thrust rapidly into the solenoid and later rapidly removed. The bulb 1. No change in, Dims out 2. Brightens in, No change out 3. Dims in, Brightens out 4. No change in, Brightens out 5. Dims in, Dims out 6. Brightens in, Brightens out 7. Brightens in, Dims out 8. Remains brighter the entire time 9. No change in or out 10. Dims in, No change out 006 10.0 poi

The correct option is;

3. Dims in, Brightens out

Explanation:

Inserting the iron rod into the current carrying solenoid causes the inductance and hence impedance to increase. Increasing impedance increases the circuit total resistance to current flow, thereby reducing the amount of  electric current, which causes the light bulb to dim as the iron rod enters the solenoid.

The reverse happens as the iron rod is rapidly removed.

The inductive reactance is given by;

Inductive reactance, = 2·π·f·L

Where;

L = Inductance (Henries)

ω = Angular velocity in radians

f = Frequency

current, I is given by

Therefore

Iron rod entering the solenoid increases inductance, decreases the current and the bulb is dimmer.

  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