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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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Using the following thermochemical equation, determine the amount of heat produced from the combustion of 24.3 g benzene (C6H6).

Anna007 [38]

Answer:

ΔH = -976.5 kJ

Explanation:

For the reaction given, there are 2 moles of benzene (C6H6). The heat of this reaction is -6278 kJ, which means that the combustion of 2 moles of benzene will lose 6278 kJ of heat. It is an exothermic reaction.

The value of ΔH, the enthalpy, is a way of measurement of the heat, and it depends on the quantity of the matter (number of moles).

So, 24.3 g of benzene has :

n = mass/ molar mass

n = 24.3/78.11

n = 0.311 moles

2 moles ------------ -6278 kJ

0.311 moles ----------- x

By a simple direct three rule:

2x = -1953.08

x = -976.5 kJ

3 0

2 years ago

If the kinetic energy of a particle is equal to twice its rest mass, what is the velocity of the particle? Determine if relativi

ivann1987 [24]

Answer:

The velocity of the particle is 2 m/s,

Explanation:

Kinetic energy is defined as energy of the body due to its motion. It is given by :

$K.E=\frac{1}{2}mv^2$

Where :

m = mass of the object

v = velocity of the object

We have , particle with mass m and its kinetic energy is twice its mass.

$K.E=2m$

$2m=\frac{1}{2}mv^2$

$v^2=\frac{4}{1}$

$v=2$

And unit of velocity are m/s , so the velocity of the particle is 2 m/s.

8 0

2 years ago

For the reaction 2N2O5(g) <---> 4NO2(g) + O2(g), the following data were colected:

KonstantinChe [14]

Answer:

a) The reaction is first order, that is, order 1. Option C is correct.

b) The half life of the reaction is 23 minutes. Option B is correct

c) The initial rate of production of NO2 for this reaction is approximately = (3.7 × 10⁻⁴) M/min. Option has been cut off.

Explanation:

First of, we try to obtain the order of the reaction from the data provided.

t (minutes) [N2O5] (mol/L)

0 1.24x10-2

10 0.92x10-2

20 0.68x10-2

30 0.50x10-2

40 0.37x10-2

50 0.28x10-2

70 0.15x10-2

Using a trial and error mode, we try to obtain the order of the reaction. But let's define some terms.

C₀ = Initial concentration of the reactant

C = concentration of the reactant at any time.

k = rate constant

t = time since the reaction started

T(1/2) = half life

We Start from the first guess of zero order.

For a zero order reaction, the general equation is

C₀ - C = kt

k = (C₀ - C)/t

If the reaction is indeed a zero order reaction, the value of k we will obtain will be the same all through the set of data provided.

C₀ = 0.0124 M

At t = 10 minutes, C = 0.0092 M

k = (0.0124 - 0.0092)/10 = 0.00032 M/min

At t = 20 minutes, C = 0.0068 M

k = (0.0124 - 0.0068)/20 = 0.00028 M/min

At t = 30 minutes, C = 0.0050 M

k = (0.0124 - 0.005)/30 = 0.00024 M/min

It's evident the value of k isn't the same for the first 3 trials, hence, the reaction isn't a zero order reaction.

We try first order next, for first order reaction

In (C₀/C) = kt

k = [In (C₀/C)]/t

C₀ = 0.0124 M

At t = 10 minutes, C = 0.0092 M

k = [In (0.0124/0.0092)]/10 = 0.0298 /min

At t = 20 minutes, C = 0.0068 M

k = 0.030 /min

At t = 30 minutes, C = 0.0050 M

k = 0.0303

At t = 40 minutes

k = 0.0302 /min

At t = 50 minutes,

k = 0.0298 /min

At t = 60 minutes,

k = 0.031 /min

This shows that the reaction is indeed first order because all the answers obtained hover around the same value.

The rate constant to be taken will be the average of them all.

Average k = 0.0302 /min.

b) The half life of a first order reaction is related to the rate constant through this relation

T(1/2) = (In 2)/k

T(1/2) = (In 2)/0.0302

T(1/2) = 22.95 minutes = 23 minutes.

c) The initial rate of production of the product at the start of the reaction

Rate = kC (first order)

At the start of the reaction C = C₀ = 0.0124M and k = 0.0302 /min

Rate = 0.0302 × 0.0124 = 0.000374 M/min = (3.74 × 10⁻⁴) M/min

3 0

2 years ago

For the reaction n2(g) + 2h2(g) â n2h4(l), if the percent yield for this reaction is 77.5%, what is the actual mass of hydrazine

Rudiy27

First calculate the moles of N2 and H2 reacted.

moles N2 = 27.7 g / (28 g/mol) = 0.9893 mol

moles H2 = 4.45 g / (2 g/mol) = 2.225 mol

We can see that N2 is the limiting reactant, therefore we base our calculation from that.

Calculating for mass of N2H4 formed:

mass N2H4 = 0.9893 mol N2 * (1 mole N2H4 / 1 mole N2) * 32 g / mol * 0.775

<span>mass N2H4 = 24.53 grams</span>

7 0

2 years ago

Determine the number of atoms in 1.85 ml of mercury. (the density of mercury is 13.5 g/ml.)

kati45 [8]

First step is to get the mass of the mercury:
Pressure = mass/volume
mass = pressure x volume = 13.5 x 1.85 = 24.975 gm

Second step is to calculate the number of moles in 24.975 gm:
From the periodic table, the molar mass of mercury is 200.59
mass = number of moles x molar mass
number of moles = 24.975 / 200.59 = 0.1245 mole

Last step is to get the number of atoms:
Each mole contains 6.02 x 10^23 atoms
number of atoms = 0.1245 x 6.02 x 10^23 = 7.4949 x 10^22 atoms

7 0

2 years ago