Suppose you want to to double a copper wire's resistance. to what temperature, in degrees celsius, must you raise it if it is originally at 24°c, neglecting any changes in dimensions? suppose you want to to double a copper wire's resistance. to what temperature, in degrees celsius, must you raise it if it is originally at 24°c, neglecting any changes in dimensions?

T=280.41 °C

Explanation:

Given that

At T= 24°C Resistance =Ro

Lets take at temperature T resistance is 2Ro

We know that resistance R given as

R= Ro(1+αΔT)

R-Ro=Ro αΔT

For copper wire

α(coefficient of Resistance) = 3.9 x 10⁻³ /°C

Given that at temperature T

R= 2Ro

Now by putting the values

R-Ro=Ro αΔT

2Ro-Ro=Ro αΔT

1 = αΔT

1 = 3.9 x 10⁻³ x ΔT

ΔT = 256.41 °C

T- 24 = 256.41 °C

T=280.41 °C

So the final temperature is 280.41 °C.

  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

1/0.14nm = 10^10/1.4 a lot