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Physics, 25.03.2020 23:33 jphan19

The total electric flux through a closed cylindrical (length = 1.2 m, diameter = 0.20 m) surface is equal to –5.0 N ⋅ m2/C. Determine the net charge within the cylinder.

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Answer from: allenpaietonp9v8sv

The net charge within the cylinder is (-44.25 pC)

Explanation:

Given that,

Total electric flux, \phi=-5\ N-m^2/C

Length of the cylinder, l = 1.2 m

Diameter of the cylinder, d = 0.2 m              

We know that the Gauss's law of electrostatics gives the relation between electric flux and the net charge. It is given by :

\phi=\dfrac{q}{\epsilon_o}

q is net charge within the cylinder

q=\phi\times \epsilon_o\\\\q=-5\times 8.85\times 10^{-12}\\\\q=-44.25\times 10^{-12}\ C\\\\q=-44.25\ pC

So, the net charge within the cylinder is (-44.25 pC). Hence, this is the required solution.            

ansver
Answer from: Quest

it is the second option.

ansver
Answer from: Quest

  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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