Please answer any you may know, will highly appreciate it, Thanks! Calculate the

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Please answer any you may know, will highly appreciate it, Thanks!

Calculate the number of molecules in a deep breath of air whose volume is 2.35 L at body temperature, 37 C, and a pressure of 730 torr molecules Submit My Answers Give Up Incorrect; Try Again; 4 attempts remaining Part B The adult blue whale has a lung capacity of 5.0 x 10L. Calculate the mass of air (assume an average molar mass 28.98 g/mol) contained in an adult blue whale's lungs at 0.2 C and 1.04 atm, assuming the air behaves ideally Express your answer using two significant figures. kg If the pressure exerted by ozone, Os, in the stratosphere is 3.0 x 10 Express your answer using two significant figures. atm and the temperature is 245 K, how many ozone molecules are in a liter? molecules Submit My Answers Give Up Incorrect; Try Again; 4 attempts remaining t B Carbon dioxide makes up approximately 0.04% o Earth's atmosphere. If you collect a 24 L sample from the atmosphere at sea level 1.00 atm on a warm day 29 d ) how many C )2 molecules are in your sample? Express your answer using two significant figures. molecules

Explanation / Answer

Solution

Given data

Volume = 2.35 L

Body temperature = 37 C +273 = 310 K

Pressure = 730 torr * 1atm / 760 torr = 0.9605 atm

Molecules in deep breath of air = ?

Here we need to use the ideal gas equation to calculate the moles of the air and then convert moles to molecules of air

Formula

PV= nRT

Where P= pressure in atm , V= volume in L, R= 0.08206 L atm per K.mol, n= moles , T= temperature in K

Now lets put the values in the formula

0.9605 atm * 2.35 L = n * 0.08206 L atm per K.mol * 310 K

2.2572 L atm = n *25.44 L atm per mol

2.2572 L atm / 25.44 L atm per mol = n

0.0887 mol air = n

Now using the moles lets calculate the molecules of air

1 mole = 6.022*1023 molecules

0.0887 mol = ? molecules

0.0887 mol * 6.022*1023 molecules / 1 mol = 5.3*1022 molecules

Therefore moles of air present in deep breath = 5.3*1022 molecules

Part B solution

Given data

Volume = 5.0*103 L

Temperature = 0.2 C +273 = 273.2 K

Pressure = 1.04 atm

Molar mass of air = 28.98 g /mol

Mass of air = ?

Here lets calculate the moles of air using the ideal gas law formula

PV= nRT

Where P= pressure in atm , V= volume in L, R= 0.08206 L atm per K.mol, n= moles , T= temperature in K

Now lets put the values in the formula

1.04 atm * 5.0*103 L = n * 0.08206 L atm per K.mol * 273.2 K

5200 L.atm = n *22.42 L atm per mol

5200 L atm / 22.42 L atm per mol = n

232 mol air = n

Now lets convert moles of Air to its mass

Mass = moles * molar mass

Mass of air = 232 mol * 28.98 g per mol

                   = 6723 g

Lets convert gram to kg

6723 g * 1 kg / 1000 g = 6.723 kg

So rounded to 6.7 kg using 2 sig fig

Part c

Given data

Pressure = 3.0*10-3 atm

Volume = 1 L

Temperature =245 K

Lets calculate the moles of O3 using the ideal gas law formula

PV= nRT

Where P= pressure in atm , V= volume in L, R= 0.08206 L atm per K.mol, n= moles , T= temperature in K

Now lets put the values in the formula

3.0*10-3 atm * 1 L = n * 0.08206 L atm per K.mol * 245 K

0.03 L.atm 1 L / 0.08206 L atm per K.mol * 245 K = n

0.00015 mol O3 = n

Now lets convert moles of O3 to its molecules

1 mole = 6.022*1023 molecules

0.00015 mol = ? molecules

0.00015 mol * 6.022*1023 molecules / 1 mol = 9.0*1019 molecules of O3

Therefore molecules of O3 = 9.0*1019 molecules of O3

Part B calculating molecules of CO2

Given data

% = 0.04 %

Volume = 2.4 L

Pressure = 1 atm

Temperature = 29 C +273 = 302 K

Lets calculate moles of CO2 using the ideal gas law formula

PV= nRT

Where P= pressure in atm , V= volume in L, R= 0.08206 L atm per K.mol, n= moles , T= temperature in K

Now lets put the values in the formula

1 atm * 2.4 L = n * 0.08206 L atm per K.mol * 302 K

2.4 L atm / 24.782 L atm per mol = n

0.09684 mol atmosphere gas = n

Now lets calculate moles of the CO2 gas using its percentage

0.09684 mol atmosphere gas * 0.04 % /100 % = 3.87*10-5 mol CO2

Now lets convert moles of CO2 to its molecules

3.87*10-5 mol CO2 * 6.022*1023 molecules / 1 mol = 2.3*1019 molecules of CO2

so molecules of CO2 in the atmosphere = 2.3*1019 molecules of CO2

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