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Marina CMI [18]
4 years ago
15

Will an empty balloon have precisely the same apparent weight on a scale as a balloon filled with air? Explain.

Chemistry
2 answers:
hram777 [196]4 years ago
4 0
The notion <span>an empty balloon have precisely the same apparent weight on a scale as a balloon filled with air depends on the diameter of the balloon. The weight of the balloon  filled with air is equal to the mass of the balloon and the mass of the air inside. The mass of air inside is equal to the density of air multiplied by the volume of the balloon. If the balloon is large, then the two masses are equal whereas if not, the mass of air inside the inflation is neglible</span>
pentagon [3]4 years ago
3 0
<span>They would not, but only for this reason: In the the full balloon, slightly more molecules of air are pressing downwards on the same position. The difference lies in the weight of the volume of the inflated balloon that does not lie in space above the deflated balloon.</span>
An air filled balloon is heavier than the empty one.
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Question 4 (1 point)
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In an endothermic reaction products are <u>HIGHER </u>than reactants in potential energy and <u>LESS </u>stable.

Explanation:

Energy is input into the reaction in an endothermic reaction. This means the products are of a higher energy level than the reactants. Therefore the reaction increases Gibb's free energy and reduces entropy. Remember in thermodynamic stability involves an increase in entropy and a decrease in Gibbs free energy. Therefore the products are less stable than the reactants. This is why endothermic reactions do not occur spontaneously like exothermic reactions.

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3 years ago
How does the density of a gas depend on the molar mass of the gas?
AlladinOne [14]

Answer:

The density of the ideal gas is directly proportional to its molar mass.

Explanation:

Density is a scalar quantity that is denoted by the symbol ρ (rho). It is defined as the ratio of the mass (m) of the given sample and the total volume (V) of the sample.

\rho = \frac{m}{V}                          ......equation (1)

According to the ideal gas law for ideal gas:

PV = nRT                                       ......equation (2)

Here, V is the volume of gas, P is the pressure of gas, T is the absolute temperature, R is Gas constant and n is the number of moles of gas

As we know,

The number of moles: n = \frac{m}{M}

where m is the given mass of gas and M is the molar mass of the gas

So equation (2) can be written as:

PV = \frac{m}{M}RT

⇒ PM= \frac{m}{V} RT

⇒ \frac{PM}{RT}= \frac{m}{V}             ......equation (3)

Now from equation (1) and (3), we get

\frac{PM}{RT}= \frac{m}{V} = \rho  

⇒ Density of an ideal gas: \rho = \frac{PM}{RT}  

⇒ <em>Density of an ideal gas: ρ ∝ molar mass of gas: M</em>

<u>Therefore, the density of the ideal gas is directly proportional to its molar mass. </u>

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