Capacitors C1 and C2 are connected in serise to a battery of variable voltage V. The capacitance of C2 is 5uF. The voltage across C2 is V2. You measure V2 as V changes. When the graph of V2 versus V is plotted, you find that it is a straight line with the slope .45.
What is the capacitance of C1?

Explanation:

In this problem, the capacitors C1 and C2 are connected in series to the battery.

The voltage of the battery, which is variable, is V.

The voltge across the capacitor C2 is V2: the graph of V2 versus V shows that a straight line with a slope of 0.45.

This means that the relationship between V and V2 is

(1)

So, 45% of the voltage is on the capacitor C2; and therefore, 55% of the voltage must be on capacitor C1:

(2)

When two capacitors are connected in series, they have the same charge on their plates, so

Using the relationship between charge (Q), capacitance (C) and voltage (V) for a capacitor:

We can rewrite the equation as

Or

Substituting (1) and (2),

And since

We find:

moving ball hits still ball. mover stops, hit ball moves. momentum transfer.

  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