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Katen [24]
3 years ago
13

How many 3-liter balloons could the 12-L helium tank pressurized to 160 atm fill? Keep in mind that an "exhausted" helium tank i

s not empty. In other words, once the gas inside the tank reaches atmospheric pressure, it will no longer be able to fill balloons. Express the number of balloons to three significant figures.
Chemistry
1 answer:
vaieri [72.5K]3 years ago
7 0

Answer:

The 12L helium tank pressurized to 160 atm will fill <em>636 </em>3-liter balloons

Explanation:

It is possible to answer this question using Boyle's law:

P_1V_1=P_2V_2

Where P₁ is the pressure of the tank (160atm), V₁ is the volume of the tank (12L), P₂ is the pressure of the balloons (1atm, atmospheric pressure) And V₂ is the volume this gas will occupy at 1 atm, thus:

160atm×12L = 1atm×V₂

V₂ = 1920L

As the tank will never be empty, the volume of the gas able to fill balloons is the total volume minus 12L, thus the volume of helium able to fill balloons is:

1920L - 12L = 1908L

1908L will fill:

1908L×\frac{1balloon}{3L} = <em>636 balloons</em>

<em></em>

I hope it helps!

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A 1.8 g sample of octane C8H18 was burned in a bomb calorimeter and the temperature of 100 g of water increased from 21.36 C to
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Answer:

HEAT OF COMBUSTION PER GRAM OF OCTANE IS 1723.08 J OR 1.72 KJ/G OF HEAT

HEAT OFF COMBUSTION PER MOLE OF OCTANE IS 196.4 KJ/ MOL OF HEAT

Explanation:

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

In other words, 3101.56 J of heat was evolved from the reaction of 1.8 g octane with water.

Heat of combustion of octane per gram:

1.8 g of octane produces 3101.56 J of heat

1 g of octane will produce ( 3101.56 * 1 / 1.8)

= 1723.08 J of heat

So, heat of combustion of octane per gram is 1723.08 J

Heat of combustion per mole:

1.8 g of octane produces 3101.56 J of heat

1 mole of octane will produce X J of heat

1 mole of octane = 114 g/ mol of octane

So we have:

1.8 g of octane = 3101.56 J

114 g of octane = (3101.56 * 114 / 1.8) J of heat

= 196 432.13 J

= 196. 4 kJ of heat

The heat of combustion of octane per mole is 196.4 kJ /mol.

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

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