A wire of Nichrome (a nickel– chromium– iron alloy commonly used in heating elements) is 1.0 m long and 1.0 mm² in cross-sectional area. It carries a current of 4.0 A when a 2.0 V potential difference is applied between its ends. Calculate the conductivity σ of Nichrome.

The conductivity of Nichrome is .

Explanation:

Given:

Potential difference (V) = 2.0 V

Current flowing (I) = 4.0 A

Length of wire (L) = 1.0 m

Area of cross section of wire (A) = 1.0 mm² = 1 × 10⁻⁶ m² [1 mm² = 10⁻⁶ m²]

We know, from Ohm's law, that the ratio of voltage and current is always a constant and equal to the resistance of the resistor. Therefore, the resistance of the nichrome wire is given as:

Now, resistance of the nichrome wire in terms of its resistivity, length and area of cross section is given as:

Where,

Now, plug in all the values given and solve for . This gives,

Now, conductivity of a material is the reciprocal of its resistivity. Therefore, the conductivity of Nichrome is given as:

Conductivity is measured in Siemens per meter (S/m)

Therefore, the conductivity of Nichrome is .

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz