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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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A student prepared an unknown sample by making a dilute solution of the unknown sample. The dilute sample was prepared by adding

telo118 [61]

Answer:

0.27 mM

Explanation:

According to the law of Lambert Beer:

A = C × E × L

Where:

A = Absorbance

C = Concentration

E = molar absorptivity

L = Step Length

If me make C the subject of the formula, we have = A / E × L

We know Absorbance and we can assume L = 1 cm, but being an unknown substance, we must know the molar Absorptivity of it, therefore, only having that value we could calculate the concentration of the mixture using the previous equation.

Then we could use the dilution ratio:

Cc * Vc = Cd * Vd

from the above formula, to fin the concentrated Concentration, we have:

Cc = Cd * Vd / Vc

we then replace the known values.

Vd = Diluted volume 25 mL

Vc = Volume concentrated 5 mL

and we would not know the diluted concentration (Cd), which is what we had to calculate in the first section. Substituting it in the previous equation, we can obtain the initial concentration.

Assuming a molar absorptivity of 5000 M-1 * cm -1 we would have:

Cd = 0.270 / 5000 * 1

= 5.4 x 10 ^ -5 M * (1000 mM / 1 M)

= 0.054 mM

and initial concentration:

Cc = 0.054 mM * (25 mL / 5 mL)

= 0.27 mM.

4 0

2 years ago

A mixture of gases containing 0.20 mol of SO2 and 0.20 mol of O2 in a 4.0 L flask reacts to form SO3. If the temperature is 25ºC

diamong [38]

Answer : The pressure in the flask after reaction complete is, 2.4 atm

Explanation :

To calculate the pressure in the flask after reaction is complete we are using ideal gas equation.

$PV=n_TRT\\\\P=(n_1+n_2)\times \frac{RT}{V}$

where,

P = final pressure in the flask = ?

R = gas constant = 0.0821 L.atm/mol.K

T = temperature = $25^oC=273+25=298K$

V = volume = 4.0 L

$n_1$ = moles of $SO_2$ = 0.20 mol

$n_2$ = moles of $O_2$ = 0.20 mol

Now put all the given values in the above expression, we get:

$P=(0.20+0.20)mol\times \frac{(0.0821L.atm/mol.K)\times (298K)}{4.0L}$

$P=2.4atm$

Thus, the pressure in the flask after reaction complete is, 2.4 atm

5 0

2 years ago

The density of a gaseous compound is 1.623 g/l at stp determine the molar mass of the compound

victus00 [196]

<span>The molar mass of the compound is 36.355 g/mol. This is calculated by knowing that 1 mol of gas fills 22.4 L of volume, so 1.623 g/L = X g/mol * 1/22.4 mol/L -> 1.623 g/L * 22.4 L/mol = X g/mol -> 36.355 g/mol = X g/mol</span>

4 0

2 years ago

The blue color in some fireworks occurs when copper(I) chloride is heated to approximately 1500 K and emits blue light of wavele

Art [367]

Given that,

Temperature, T = 1500 K

Wavelength, $\lambda=4.5\times 10^2\ nm$