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2 years ago

Explain why a propane torch is lit inside a hot air balloon during preflight preparations. Which gas law applies?

Chemistry

2 answers:

xenn [34]2 years ago

7 0

Answer: The gas law which applies here is :Charle's Law

Explanation:

The main principle behind this is that, warm air rises in the cooler air.Which implies that hot air is less denser than the cooler air. Due to this property hot air rises up in the sky.

Density decrease with increase in volume due to this hot air balloon rises up in the air

Also the volume occupied by the gas increase with increase in temperature at constant pressure known as Charle's law.

So, during preflight the propane torch is lit to warm up the air present inside the balloon.

Olegator [25]2 years ago

5 0

Propane torch is lit inside a hot air balloon during pre-flight preparation because the heat from the touch is needed to heat the cold air inside the balloon, so that the air will expand and become less dense and rise, thus providing a lift for the balloon. This is line with charle's law, which states that, the volume of a fixed mass of ideal gas is directly proportional to the absolute temperature. This law implies that, as the temperature of the air inside the balloon increase, the volume of the balloon also increases.

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If a certain gas occupies a volume of 13 L when the applied pressure is 6.5 atm , find the pressure when the gas occupies a volu

Sav [38]

We should apply Boyle's Law here given initial pressure, initial volume and final volume.

P1V1= P2V2
(6.5 atm) (13 L) = P2 (3.3 L)

Solve for P2 on your calculator and that should get you to the answer.

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2 years ago

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As Jesse's hybrid accelerates to pass a car on the highway, he notices that his gas mileage drops from 40 miles per gallon to 15

e-lub [12.9K]

The most likely explanation for this observation is C. his car has turned on more pistons to provide the extra energy needed to accelerate.

When cruising, hybrid cars are able to employ electrical energy to drive the car. Moreover, even if a vehicle is not a hybrid, a greater amount of fuel is consumed when one accelerates because the vehicle has to generate a force larger than the force of air resistance in order for it to accelerate. This increased demand of force reduces the vehicle's fuel economy.

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2 years ago

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Which of the following is a correctly written chemical equation that demonstrates the conservation of mass?

Fantom [35]

Answer:

Option D is correct.

H₂O + CO₂ → H₂CO₃

Explanation:

First of all we will get to know what law of conservation of mass states.

According to this law, mass can neither be created nor destroyed in a chemical equation.

This law was given by French chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

Example:

6CO₂ + 6H₂O + energy → C₆H₁₂O₆ + 6O₂

there are six carbon atoms, eighteen oxygen atoms and twelve hydrogen atoms on the both side of equation so this reaction followed the law of conservation of mass.

Now we will apply this law to given chemical equations:

A) H₂ + O₂ → H₂O

There are two hydrogen and two oxygen atoms present on left side while on right side only one oxygen and two hydrogen atoms are present so mass in not conserved. This equation not follow the law of conservation of mass.

B) Mg + HCl → H₂ + MgCl₂

In this equation one Mg, one H and one Cl atoms are present on left side while on right side two hydrogen, one Mg and two chlorine atoms are present. This equation also not follow the law of conservation of mass.

C) KClO₃ → KCl + O₂

There are one K, one Cl and three O atoms are present on left side of chemical equation while on right side one K one Cl and two oxygen atoms are present. This equation also not following the law of conservation of mass.

D) H₂O + CO₂ → H₂CO₃

There are two hydrogen, one carbon and three oxygen atoms are present on both side of equation thus, mass remain conserved. Thus is correct option.

6 0

2 years ago

Which mass of gas would occupy a volume of 3dm3 at 25°C and 1 atmosphere pressure?

balu736 [363]

Answer:

C 8.09 SO2 gas

Explanation:

As we have the volume (3dm³ = 3L), temperature (25°C + 273 = 298K), and pressure (1atm), we can solve to moles of gas using:

PV = nRT

PV / RT = n

1atm*3L / 0.082atmL/molK*298K =¨

0.123 moles of gas you have.

Now, to convert these moles to mass we use molar mass (32g/mol for O2, 28g/mol for N2, 64g/mol for SO2, and 44g/mol for CO2).

Mass of 0.123 moles of these gases is:

O2 = 0.123 moles * 32g/mol = 3.94g of O2. A is wrong

N2 = 0.123 moles * 28g/mol = 3.4g of N2. B is wrong

SO2 = 0.123 moles * 64.1g/mol = 7.9g of SO2≈ 8.09g of SO2, C is possible

CO2 = 0.123 moles * 44g/mol = 5.4g of CO2. D is wrong

Right answer is:

8 0

2 years ago

A solution of 20.0 g of which hydrated salt dissolved in 200 g H2O will have the lowest freezing point? (A) CuSO4 • 5 H2O (M = 2

Andrews [41]

Answer:

(D) Na₂SO₄•10H₂O (M = 286).

Explanation:

The depression in freezing point of water by adding a solute is determined using the relation:

ΔTf = i.Kf.m,

Where, <em>ΔTf </em>is the depression in freezing point of water.

<em>i</em> is van't Hoff factor.

<em>Kf </em>is the molal depression constant.

<em>m</em> is the molality of the solute.

Since, Kf and m is constant for all the mentioned salts. So, the depression in freezing point depends strongly on the van't Hoff factor (i).

van't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.

(A) CuSO₄•5H₂O:

CuSO₄ is dissociated to Cu⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(B) NiSO₄•6H₂O:

NiSO₄ is dissociated to Ni⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(C) MgSO₄•7H₂O:

MgSO₄ is dissociated to Mg⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(D) Na₂SO₄•10H₂O:

Na₂SO₄ is dissociated to 2 Na⁺ and SO₄²⁻.

So, i = dissociated ions/no. of particles = 3/1 = 3.

∴ The salt with the high (i) value is Na₂SO₄•10H₂O.

So, the highest ΔTf resulted by adding Na₂SO₄•10H₂O salt.

4 0

2 years ago