1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
loris [4]
3 years ago
7

At a festival, spherical balloons with a radius of 170.cm are to be inflated with hot air and released. The air at the festival

will have a temperature of 25°C and must be heated to 100°C to make the balloons float. 1.00kg of butane C4H10 fuel are available to be burned to heat the air. Calculate the maximum number of balloons that can be inflated with hot air.
Chemistry
1 answer:
MissTica3 years ago
7 0

Answer:

The correct answer is 1.5 balloons.

Explanation:

Based on the given question, the radius of the spherical balloons is 170 cm. Therefore, the volume of each balloon will be,  

= 4/3 × π × (170)³ cm³

= 20.5176 × 10⁶ cm³  

= 20.5176 m³

The density of air at 100 degree C s 0.946 Kg m⁻³

The mass of air in each balloon can be calculated by using the formula,  

Mass = density × volume

Mass = 0.946 Kg m⁻³ × 20.5176 m³

Mass = 19.410 Kg

The heat energy, that is, required to bring the air from 25 degree C to 100 degree C will be,  

= 19.410 × 10³ g × (100 -25) degree C × 1.009 J/g degree C

= 14.68 × 10⁵ J

The concentration of butane given is 1.00 Kg or 1000 g

The molecular weight of butane is 58.12 g per mole

The moles or n can be calculated by using the formula,  

n = mass / mol.wt

n = 1000 g/58.12 g/mol = 17.20 mol

The formation of enthalpy of butane at 25 degree C is 125.7 × 10³ J/mol. The evolution of heat energy that take place at the time of burning 17.20 moles of butane is,  

= 125.7 × 10³ J/mol × 17.20 mol

= 2.16 × 10⁶ J

The number of balloons that can be inflated with hot air is,  

= 2.16 × 10⁶ J / 14.68 × 10⁵ J per each balloon

= 1.5 balloons

Hence, maximum of 1.5 balloons can be inflated.  

You might be interested in
Eq-43 which type of fire extinguisher should be onboard a vessel with a permanently installed fuel tank
liq [111]
I think a type B fire extinguisher should be on board a vessel with a permanently installed fuel tank. 
All vessels are required to have a type B fire extinguisher on board if one or more of the following conditions exists. That is; Inboard engine, vessel length of 26 feet or longer, enclosed living spaces, closed storage compartments in which flammable or combustible materials may be stored, permanently installed fuel tanks, and also closed compartments where portable fuel tanks may be stored. 
7 0
3 years ago
2 FONS
stiv31 [10]

Answer:

C. It decreases by a factor of 4

Explanation:

F1 = kq1*q2/r²

F2 = kq1*q2/(2r)² = kq1*q2/(4r²) = kq1*q2/(r²*4)  = F1/4

7 0
3 years ago
A gaseous mixture of O2 and N2 contains 37.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture if the tot
kondor19780726 [428]

Answer: The partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

Explanation:

mass of nitrogen = 37.8 g

mass of oxygen = (100-37.8) g = 62.2 g

Using the equation given by Raoult's law, we get:

p_A=\chi_A\times P_T

p_{O_2} = partial pressure of O_2 = ?

\chi_{O_2} = mole fraction of O_2=\frac{\text{Moles of }O_2}{\text{Total moles}}

P_{T} = total pressure of mixture  = 525 mmHg

{\text{Moles of }O_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{62.2g}{32g/mol}=1.94moles

{\text{Moles of }N_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{37.8g}{28g/mol}=1.35moles

Total moles = 1.94 + 1.35 = 3.29 moles

\chi_{O_2}=\frac{1.94}{3.29}=0.59

p_{O_2}=\chi_{O_2}\times P_T=0.59\times 525=310mmHg

Thus the partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

7 0
3 years ago
Add the polynomials (7x^3-2x^2-12)+(3x^3-8x^2+10x)
laiz [17]

Answer:

10 x ^3 − 10 x ^2 + 10 x − 12

Explanation:

4 0
2 years ago
What is the percent yield of O2 if 10.2 g of O2 is produced from the decomposition of 17.0 g of H2O?
yawa3891 [41]

The balanced chemical reaction will be:

2H2O = 2H2 + O2

<span>We are given the amount of water used in the decomposition reaction. This will be our starting point.</span>

<span>17.0 g H2O</span> (1 mol  H2O/ 18.02 g H2O) (1 mol O2/2 mol <span>H2O</span>) ( 32.00 g O2/1mol O2) = 15.09 g O2

Percent yield = actual yield / theoretical yield x 100

<span>Percent yield =10.2 g / 15.09  g x 100</span>

Percent yield = 67.58%

5 0
3 years ago
Read 2 more answers
Other questions:
  • Give the product of the reaction of excess benzene (2 equivalents) with dichloromethane and alcl3.
    6·1 answer
  • Which of the following for an object should be known to calculate its density
    8·1 answer
  • A student adds 4.00g of dry ice (solid CO2) to an empty balloon. What will be the volume of the balloon at STP after all the dry
    15·1 answer
  • What is the formula for aluminum sulfate?
    13·1 answer
  • Aqueous carbonic acid H2CL3decomposes into carbon dioxide gas and liquid water . Write a balanced chemical equation for this rea
    15·2 answers
  • The molecular mass of the compound is 132 amu. What is the molecular formula?
    8·1 answer
  • Which type of decay does not involve the decaying atom's atomic number to change?
    9·1 answer
  • 200 feet in 30 seconds is what speed?
    15·2 answers
  • What would the chemical formula be of a compound that has Hydrogen and Carbon in it?
    5·2 answers
  • Explain any three ways of protecting our water bodies
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!