g The current through a 10 m long wire has a current density of 4 cross times 10 to the power of 6 space open parentheses bevelled A over m squared close parentheses. The wire conductivity is 2 cross times 10 to the power of 7 space open parentheses bevelled S over m close parentheses. Find the voltage drop across the wire. (Answer with the numeric value, don't write the unit V)

The value is

Explanation:

From the question we are told that

The length of the wire is

The current density is

The conductivity is

Generally conductivity is mathematically represented as

Here R is the resistance which is mathematically represented as

Here I is the current which is mathematically represented as

So

And

=>

=>

=>

=>

  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

i think they are all correct

explanation:

they seem good