A tire is filled with air at 22oC to a gauge pressure of 240 kPa. After driving for some time, if the temperature of air inside the tire is 45oC, what fraction of the original volume of air must be removed to maintain the pressure at 240 kPa?

7.8% of the original volume.

Explanation:

From the given information:

Temperature = 22° C = 273 + 22 = 295° C

Pressure = 240 kPa

Temperature = 45° C

At initial temperature and pressure:

Using the ideal gas equation:

making V_1 (initial volume) the subject:

Provided the pressure maintained its rate at 240 kPa, when the temperature reached 45° C, then:

the final volume can be computed as:

Now, the change in the volume ΔV =  V₂ - V₁

∴

The required fraction of the volume of air to keep up the pressure at (240) kPa can be computed as:

= 7.8% of the original volume.

  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

massxgravity x heght=weightxheight= pe

4900=490xh

h=10