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White raven [17]
2 years ago
15

The balloon Spirit of Freedom, flown around the world by American Steve Fossett in 2002, contained 550,000 cubic feet of helium.

How much PΔV work was done by the balloon on the surrounding atmosphere, assuming atmospheric pressure was 1.00 atm while the balloon was being inflated?
Chemistry
1 answer:
Elodia [21]2 years ago
3 0
Assuming the balloon initially has volume of 0 when deflated, the total P(deltaV) = (1.00 atm)(550,000 ft^3) = 550,000 atm-ft^3. To convert into work units, we can first convert ft^3 to L:
(550,000 atm-ft^3)(1 m/3.28 ft)^3
= (15,586.2 atm-m^3)
Then we convert to L:
(15,586.2 atm-m^3)(1000 L/m^3)
= 15,586,200 atm-L
Then we convert to J:
(15,586,200 atm-L)(101.325 J / 1 atm-L)
= 1.579 x 10^9 J
You might be interested in
How many grams of NH3 can be prepared from the synthesis of 77.3 grams of nitrogen and 14.2 grams of hydrogen gas?
lbvjy [14]

Answer:

80.41 g

Explanation:

Data Given:

Mass of Nitrogen (N₂) = 77.3 g

Mass of Hydrogen (H₂) = 14.2 g

many grams of NH₃ = ?

Solution:

First we look at the balanced synthesis reaction

              N₂   +    3 H₂  ------—> 2 NH₃

             1 mol      3 mol

As 1 mole of Nitrogen react with 3 mole of hydrogen

Convert moles to mass

molar mass of N₂ = 2(14) = 28 g/mol

molar mass of H₂ = 2(1) + 2 g/mol

Now

                     N₂             +           3 H₂        ------—>      2 NH₃

             1 mol (28 g/mol)     3 mol(2g/mol)

                    28 g                        6 g

28 grams of N₂ react with 6 g of H₂  

So

if 28 grams of N₂ produces 6 g of H₂  so how many grams of N₂ will react with 14.2 g of H₂.

Apply Unity Formula

                 28 g of N₂ ≅ 6 g of H₂

                 X g of N₂ ≅ 14.2 g of H₂

Do cross multiply

                X g of N₂ = 28 g x 14.2 g / 6 g

                X g of N₂ = 66.3 g

As we have given with 77. 3 g of N₂ but from this calculation we come to know that 66.3 g will react with 14.2 g of hydrogen and the remaining 10 g N₂ will be in excess

So, Hydrogen is limiting reactant in this reaction and the amount of NH₃ depends on the amount of hydrogen.

Now

To find mass of NH₃ we will do following calculation

Look at the reaction

As we Know

                     N₂             +           3 H₂        ------—>      2 NH₃

                                                   6 g                            2 mol

So, 6 g of hydrogen gives 2 moles of NH₃, then how many moles of NH₃ will be produce by 14.2 g

Apply Unity Formula

                 6 g of H₂ ≅ 2 mol of NH₃

                14.2 g of H₂ ≅ X mol of NH₃

Do cross multiply

               X mol of NH₃= 14.2 g x 2 mol / 6 g

                X mol of NH₃ = 4.73 mol

So, 14.2 g of hydrogen gives 4.73 moles of NH₃

Now

Convert moles of NH₃ to mass

Formula will be used

        mass in grams = no. of moles x molar mass . . . . . . (2)

Molar mass of  NH₃

Molar mass of  NH₃ = 14 + 3(1)

Molar mass of  NH₃ = 14 + 3 = 17 g/mol

Put values in equation 2

        mass in grams = 4.73 mole x 17 g/mol

        mass in grams =  80.41 g

mass of NH₃=  80.41 g

3 0
2 years ago
Which of the following compounds would be most effective in lowering the melting point of ice on roads? a) CaCl2 b) NaCl C) K3PO
Mrac [35]
I'm not 100% sure on this, but I would go with C) NaCl. 
NaCl is a salt, and that is used to melt the ice on the roads. Hope this helps!
7 0
2 years ago
A 0.200 M K 2SO 4 solution is produced by ________. dilution of 250.0 mL of 1.00 M K2SO4 to 1.00 L dissolving 43.6 g of K2SO4 in
Artist 52 [7]

A 0.200 M of K2SO4 solution is produced by diluting 20.0 mL of 5.00 M K2SO4 solution to 500.0 mL.  

<u>Explanation</u>:

  •  When dealing with dilution we will use the following equation:

                              M1 V1 = M2 V2

where,

                      M1 = initial concentration

                      V1 = initial volume

                      M2 = final concentration

                      V2 = final volume

  • By diluting 20.0 mL of 5.00 M K2SO4 solution to 500.0 mL, we get

                            M1 V1 = M2 V2

     20.0 mL \times    5.00 M = M2 \times 500.0 mL

                               M2 = (20.0 mL \times    5.00 M) / 500.0 mL

                              M2 =  0.200 M.

Hence A 0.200 M of K2SO4 solution is produced by diluting 20.0 mL of 5.00 M K2SO4 solution to 500.0 mL.  

3 0
2 years ago
The metallic radius of a potassium atom is 231 pm. What is the volume of a potassium atom in cubic meters?
Sauron [17]
1 pm = 10∧-10 cm
Therefore, 230 pm is equivalent to 2.3 ×10∧-8 cm.
Atom is in the shape of a sphere,
The volume of a sphere is given by 4/3πr³
Thus, volume of the atom = 4/3π( 2.3 ×10∧-8)³
                                          = 4/3 (3.142 ×12.167×10∧-24
                                           = 5.096 ×10∧-23 cm³
but 1m³= 1000000cm³
Therefore, the volume of the atom = 5.096 ×10∧-29 m³
8 0
2 years ago
In order to reduce the exposure to organic solvents like turpentine, some art instructors recommend the students clean brushes a
kykrilka [37]
Generally speaking, organic molecules tend to dissolve in solvents that have similar physical properties. A good rule of thumb is that "like dissolves like". Meaning, polar compounds can dissolve polar compounds and nonpolar compounds can dissolve nonpolar compounds.

To apply this to the current problem, we are told that the brushes are being cleaned with vegetable oil or mineral oil. In this case, the oils are used as solvents. In order for these solvents to be effective, the compounds they are trying to dissolve must be similar in structure and properties to other oils. Therefore, vegetable oil or mineral oil will be most effective in removing oil-based paints, as these will have the similar properties needed to dissolve in the oil solvents.
6 0
2 years ago
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