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3 years ago

How many atoms of Pb are there in the compound 3Pbl2

Physics

2 answers:

mash [69]3 years ago

5 0

Answer: there are 6 atoms.

Explanation: 3 parts of Pb times 2 atoms per part = 6

good luck:)

Setler [38]3 years ago

3 0

Answer:

6 atoms

Explanation:

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A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the

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Answer:

$Q=3.9825\times 10^{-9} C$

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S= $\frac{360}{10000}=0.036 m^2$

$\Delta d=0.8 cm=0.008 m$

$\Delta V=100 V$

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

$C=\frac{\epsilon_0 S}{d}$

Capacitance of capacitor after moving plates

$C_1=\frac{\epsilon_0 S}{(d+\Delta d)}$

$V=\frac{Q}{C}$

Potential difference between plates after moving

$V=\frac{Q}{C_1}$

$V+\Delta V=\frac{Q}{C_1}$

$\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}$

$\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100$

$\frac{Q\Delta d}{\epsilon_0 S}=100$

$\epsilon_0=8.85\times 10^{-12}$

$Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}$

$Q=3.9825\times 10^{-9} C$

Hence, the charge on positive plate of capacitor= $Q=3.9825\times 10^{-9} C$

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