What is the density ρh of hot air inside the balloon? assume that this density is uniform throughout the balloon. express the density in terms of th, tc, and ρc?

Answers

Answer 1
Answer:

With the Ideal Gas formula.

Further Explanation

The ideal gas equation is an equation that presents the relationship between the pressure and volume of a gas with the temperature and the number of moles of the gas itself. This ideal gas equation is based on the laws of Boyle, Charles, and Avogadro.

This equation is generally written as

-PV = nRT

Boyle's Law (T = C)

Boyle's Law about the nature of gas was first created by Robert Boyle (1627¡1691). Boyle found experimentally that at a constant temperature, the pressure in a closed container would be inversely proportional to the volume of the gas. This means that if we press the gas and large volumes to small volumes, the gas pressure will rise, conversely if we expand the gas from small volumes to large volumes, the gas pressure will decrease. The process of pressing gas from large volumes to small volumes is called the compression process. While the process of changing the volume of gas from a small volume into a large volume is called the expansion process.

Gay Lussac Law (V = C)

The Gay Lussac Law was first coined by Joseph Louis Gay-Lussac (1778 - 1850) a chemist from France. Lussac's Gay Law states that under constant volume conditions, an increase in gas pressure will be directly proportional to an increase in gas temperature. This means that the gas in a closed container with a volume does not change (V0 = V1), so if the gas is heated from the initial temperature T0 to the final temperature T1 will cause the gas pressure also rises from the initial pressure (p0) to the final pressure (p1).

Charles Law (P = C)

Charles' ideal gas law is also called the law of constant pressure, first created by Jacques Alexandre Cesar Charles (1746 - 1823) a chemist from France. Charles's Law says "Under conditions of constant gas pressure, the volume of a gas will increase in proportion to the increase in temperature of the gas". This means that if we have a cylinder-piston system that can move up and down freely, then when the gas in the system is heated so that the temperature rises from T0 to T1, then to keep the gas pressure in the cylinder constant, the volume of gas will expand from the condition V0 to V1.

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Grade:  College

Subject:  Chemistry

keywords: Gas, the ideal gas.


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What is the correct equilibrium constant expression for the following reaction? 3A2 = 2B3 when the reaction started with the initial concentrations of A2 = 3 M and B3 = 2 M and continued until the equilibrium concentrations of A2 = 2.5 M and B3 = 2.5 M

Answers

Answer:

Kc = [B₃]²/[A₂]³  = 0.40

Explanation:

3A₂ ⇄ 2B₃  

Given at equilibrium =>  [A₂] =2.5 and [B₃] = 2.5

Kc = [B₃]²/[A₂]³ = (2.5)²/(2.5)³ = (2.5)⁻¹ = 0.40

The [Fe(H₂O)₆]³⁺ complex requires a relatively small amount of energy to promote an electron from the t2g to the eg. Based on the UV data, predict the spin of this complex.

Answers

Answer:

The spin of the complex is 5.92 B.M

Explanation:

Please see the attachments below

Final answer:

The [Fe(H₂O)₆]³⁺ complex is a high-spin complex due to the relatively small energy required to promote an electron from the t2g to the eg orbital. As such, it is reasonable to predict that it has a high-spin state with five unpaired electrons.

Explanation:

The [Fe(H₂O)₆]³⁺ complex is a type of coordination complex in which the central metal atom, Fe³⁺, is surrounded by six water molecules acting as ligands. The spin state of such a complex can be determined based on the energy required to promote an electron from the t2g to the eg.

In [Fe(H₂O)₆]³⁺, the field produced by the water ligands is relatively weak, resulting in a small crystal field splitting (Aoct <P). Given that it requires less energy for the electrons to occupy the eg orbitals than to pair up, there will be an electron in each of the five 3d orbitals before any pairing occurs. Hence, for the six d electrons on the Fe³⁺ ion in [Fe(H₂O)₆]³⁺, there should be one pair (two electrons) and four unpaired electrons.

High-spin complexes are those in which the electrons tend not to pair up because the crystal field splitting is not large enough to make it energetically favorable for them to do so. Given that the [Fe(H₂O)₆]³⁺ complex falls under the categories of high-spin complexes, it is reasonable to predict that it exhibits a high-spin state with five unpaired electrons.

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How many grams of chloride are there in 256 g of magnesium chloride?

Answers

Given :

Mass of given magnesium chloride, m = 256 g.

To Find :

How many grams of chloride are there in 256 g of magnesium chloride.

Solution :

Molecular formula of magnesium chloride is MgCl_2.

Molecular formula of MgCl_2 is, M = 95.211 g/mol .

Mass of chlorine in 1 mol of MgCl_2  is , m = 35.5 × 2 = 71 g.

So, amount of chlorine in 256 gram MgCl_2 :

m =(71* 256)/(95.211)\n\nm = 190.90\ g

Hence, this is the required solution.

Calculate the mass of the air contained in a room that measures 2.50 m × 5.50 m × 3.00 m (density of air = 1.29 g/dm3 at 25°C).

Answers

The mass of air in room as per given density is 53.2 kg.

To calculate the mass of air contained in a room, we can use the formula:

mass = density x volume

Here, the given density of air is 1.29 g/dm³ at 25°C. We can convert the dimensions of the room to decimeters (dm) by multiplying by 10:

Length = 2.50 m × 10 = 25 dm

Width = 5.50 m × 10 = 55 dm

Height = 3.00 m × 10 = 30 dm

Now, we can calculate the volume of the room by multiplying the three dimensions:

Volume = length x width x height

Volume = 25 dm x 55 dm x 30 dm

Volume = 41,250 dm³

Finally, we can use the formula to calculate the mass of air:

mass = density x volume

mass = 1.29 g/dm³ x 41,250 dm³

mass = 53,212.5 g or 53.2 kg

Therefore, the mass of air contained in the room is approximately 53.2 kg.

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Explanation:

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The covalent compounds are soluble in:a) All acids

b) All bases

c) all solvents

d) nonpolar solvents​

Answers

Answer:

d) Non-polar solvents

1) How many kJ are absorbed when 45.2 g of water at 31.3 oC is heated to 76.9 oC? 2) Calculate the total heat released in kcal when 72.1 g water at 25.2 oC is cooled to 0 oC and freezes. 3) How many kilojoules are required to heat 55,500 mg of gold with specific heat = 0.129 J/g oC is heated from 24.6 oC to 123.4 oC? 4) Calculate the heat needed in kcal to change 45.6 g of water at 100 oC to change into steam.

Answers

Answer:

1. Q = 8.66 KJ

2. Q = 7.58 Kcal

3. Q = 0.71 KJ

4. Q = 24.31 Kcal

Explanation:

1. The quantity of heat absorbed can be determined by:

       Q = mcΔθ

where: Q is the quantity of heat absorbed or released, m is the mass of the substance, c is the specific heat capacity of water = 4.2 j/g^(0)C and Δθ is the change in temperature.

           = 45.2 × 4.2  × (76.9 - 31.3)

           = 8656.704

∴       Q = 8.66 KJ

The quantity of heat absorbed is 8.66 KJ.

2. Q = mcΔθ + mL

Where L is the latent heat of fusion of ice = 334 J.

       = m(cΔθ + L)

       = 72.1(4.2 × 25.2 + 334)

   Q = 31712.464 J

       = 7579.466 calories

The total heat released is 7.58 Kcal.

3. Q = mcΔθ

      = 55.5 × 0.129 × (123.4 - 24.6)

     = 707.3586

The quantity of heat required to increase the temperature of gold is 0.71 KJ.

4. Q = mL

Where: L is the specific latent heat of vaporization = 533 calories.

     Q  = 45.6 × 533

      = 24304.8

The quantity of heat required to change water to steam is 24.31 Kcal.