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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.  

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The concentration of protein in a urine sample is calculated to be 2.77 μg/mL. What is the concentration of this solution in uni
maxonik [38]

Answer:

The concentration of this solution in units of pounds per gallon is 2.776*10^{-5} \frac{lb}{gal}

Explanation:

Units of measurement are established models for measuring different quantities. The conversion of units is the transformation of a quantity, expressed in a certain unit of measure, into an equivalent one, which may or may not be of the same system of units.

In this case, the conversion of units is carried out knowing that 1 μg are equal to 2.205*10⁻⁹ Lb and  1 mL equals 0.00022 Gallons. So

2.77 \frac{ug}{mL} = \frac{2.77 ug}{mL}

If 1 μg equals 2.205*10⁻⁹  lb, 2.77 μg how many lb equals?

lb=\frac{2.77ug*2.205*10^{-9}lb }{1ug}

lb=6.10785*10⁻⁹

So, 2.77 μg= 6.10785*10⁻⁹ lb

Then:

2.77 \frac{ug}{mL} = \frac{2.77 ug}{mL}=\frac{6.10785*10^{-9}lb }{mL} =\frac{6.10785*10^{-9}lb }{0.00022 gal} =\frac{6.10785*10^{-9}lb }{0.00022 gal}

You get:

2.77 \frac{ug}{mL} = 2.776*10^{-5} \frac{lb}{gal}

<u><em>The concentration of this solution in units of pounds per gallon is </em></u>2.776*10^{-5} \frac{lb}{gal}<u><em></em></u>

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Please help.
mojhsa [17]

Explanation: An element is represented in the form of _{A}^{Z}\textrm{X} where,

X = Symbol of the element

A = Atomic Mass of the element X

Z = Atomic Number of element X

Hence, For the element Justwondoricium,

Symbol = Jw

Atomic number = 120

Atomic Mass = 224

Now, Atomic number = Number of electrons = Number of Protons

Number of electrons = 120

Number of Protons = 120

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Number of neutrons can be calculated as we know atomic mass and number of neutrons, Putting the numbers we get

Number of neutrons = 224 - 120 = 104

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As the electron is entering the 8th shell, it belongs to the 8th period and as the last electron enters the s-orbital, it belongs to the S-block of the periodic table.

When the s-electrons are 1, it belongs to Group 1.

When the s-electrons are 2, it belongs to Group 2.

As this element has 2 s-electrons, it belongs to the Group 2.

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