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torisob [31]
3 years ago
13

Whitney's lung capacity was measured as 3.3 l at a body temperature of 37 ∘c and a pressure of 746 mmhg . how many moles of oxyg

en is in her lungs if air contains 21% oxygen?
Chemistry
1 answer:
myrzilka [38]3 years ago
4 0

the ideal gas equation is PV=nRT  
 where P=pressure  
 V=Volume  
 n=no. of moles  
 R=universal gas constant  
 T=temperature  
 The universal gas constant (R) is 0.0821 L*atm/mol*K    
 a pressure of 746 mmhg =0.98 atm= 1 atm (approx)   
 T=37 degrees Celsius =37+273=310 K (convert it to Kelvin by adding 273)    
 V=0.7 L (only getting oxygen, get 21% of 3.3L)    
 Solution:  
 (1 atm)(0.7 L)=n(0.0821 L*atm/mol*K)(310 K)  
 0.7 L*atm=n(25.451 L*atm/mol)  
 n=0.0275 mole   
 Answer:   
n=0.0275 mole of oxygen in the lungs.
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Answer:

44,901 kilo Joule heat is released when 1.663\times 10^4 g grams of ammonia is produced.

Explanation:

Moles of ammonia  gas produced :

\frac{1.663\times 10^4 g}{17 g/mol}=978.235 mol

According to reaction, when 2 moles of ammonia are produced 9.18 kilo joules of energy is also released.

So, When 978.235 moles of ammonia gas is produced the energy released will be:

\frac{-91.8 kJ}{2}\times 978.235 mol=-44,900.98 kJ\approx -44,901 kJ

(negative sign indicates that energy is released as heat)

44,901 kilo Joule heat is released when 1.663\times 10^4 g grams of ammonia is produced.

7 0
3 years ago
The combustion of 1.00 mol of glucose, C6H12O6, releases 2820 kJ of heat. If 2.0 g of glucose is burned in a calorimeter contain
DochEvi [55]

Answer:

The heat capacity of the calorimeter is 4.76 kJ/°C

Explanation:

Step 1: Data given

1.00 mol of glucose releases 2820 kJ of heat

Mass of glucose = 2.0 grams

Mass of water = 1000 grams

The temperature increases with 3.5 °C

Step 2: Calculate moles

moles glucose = mass glucose / molar mass glucose

moles glucose = 2.0 grams / 180.16 g/mol

moles glucose = 0.0111 moles

Step 3: Calculate heat produced by the combustion

Heat produced = 2820 kJ/mol * 0.0111 moles

Heat produced = 31.302 kJ = 31302 J

Step 4: Calculate heat absorbed by the water

Q = m*c*ΔT

⇒ with m = the mass of water = 1000 grams

⇒ with c = the specific heat of water = 4.184 J/g°C

⇒ with ΔT = The change in temperature = 3.5 °C

Q = 1000 * 4.184 *3.5

Q = 14644 J absorbed by the water

Step 5: Calculate heat basorbed by the calorimeter

Q = 31302 - 14644 = 16658 J absorbed by the calorimeter

Step 6: Calculate the heat capacity of the calorimeter

c= 16658 J / 3.5 °C

c = 4759 J/°C = 4.76 kJ/°C

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3 years ago
A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant?
KengaRu [80]

Answer:

d. Copper (II) sulfate

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Given data:

Mass of Al = 1.25 g

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What is limiting reactant = ?

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Chemical equation:

2Al + 3CuSO₄   →   Al₂ (SO₄)₃ + 3Cu

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Number of moles = mass/molar mass

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now we will compare the moles of reactant with product.

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                Al           :            Cu

                 2            :              3

               0.05         :            3/2×0.05 = 0.075 mol

         CuSO₄           :           Al₂ (SO₄)₃

                3             :             1

               0.02         :          1/3×0.02=0.007 mol

         CuSO₄           :            Cu

               3               :              3

               0.02         :              0.02

Less number of moles of reactants are produced by CuSO₄ thus it will act as limiting reactant.

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