What is the approximate weight of the air inside the tire in english engineering units (tire outside diameter = 49", rim diameter-22", tire width-19" see: tire side view. ndf?

1.265 Pounds

Explanation:

Data provided:

Tire outside diameter = 49"

Rim diameter = 22"

Tire width = 19"

Now,

1" = 0.0254 m

thus,

Tire outside radius, r₁ = 49"/2 = 24.5" = 24.5 × 0.0254 = 0.6223 m

Rim radius, r₂   = 22" / 2 = 11" = 0.2794 m

Tire width, d = 19" = 0.4826 m

Now,

Volume of the tire = π ( r₁² - r₂² ) × d

on substituting the values, we get

Volume of air in the tire = π (  0.6223² - 0.2794² ) × 0.4826 = 0.46877 m³

Also,

Density of air = 1.225 kg/m³

thus,

weight of the air in the tire = Density of air × Volume air in the tire

or

weight of the air in the tire = 1.225 × 0.46877 = 0.5742 kg

also,

1 kg = 2.204 pounds

Hence,

0.5742 kg = 0.5742 × 2.204  = 1.265 Pounds

  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