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3 years ago

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Use the ideal-gas law to estimate the number of air molecules in your physics lab room, assuming all the air is N2. Assume a roo

m size of 18 ft × 18 ft × 18 ft . Assume a temperature of 20∘C.

1 answer:

stiv31 [10]3 years ago

4 0

Answer:

4.13×10²⁷ molecules of N₂ are in the room

Explanation:

ideal gases Law → P . V = n . R . T

Pressure . volume = moles . Ideal Gases Constant . T° K

T°K = T°C + 273 → 20°C + 273 = 293K

Let's determine the volume of the room:

18 ft . 18 ft . 18ft = 5832 ft³

We convert the ft³ to L → 5832 ft³ . 28.3L / 1 ft³ = 165045.6 L

1 atm . 165045.6 L = n . 0.082 L.atm/mol.K . 293K

(1 atm . 165045.6 L) / 0.082 L.atm/mol.K . 293K = n

6869.4 moles of N₂ are in the room

If we want to find out the number of molecules we multiply the moles by NA

6869.4 mol . 6.02×10²³ = 4.13×10²⁷ molecules

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The following diagrams represent mixtures of NO(g) and O2(g). These two substances react as follows: 2NO(g)+O2(g)→2NO2(g) It has

Alja [10]

This is an incomplete question, here is a complete question and an image is attached below.

The following diagrams represent mixtures of NO(g) and O₂(g). These two substances react as follows:

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

It has been determined experimentally that the rate is second order in NO and first order in O₂.

Based on this fact, which of the following mixtures will have the fastest initial rate?

The mixture (1). The mixture (2). The mixture (3).

Answer : The mixture 1 has the fastest initial rate.

Explanation :

The given chemical reaction is:

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

The rate law expression is:

$Rate=k[NO]^2[O_2]$

Now we have to determine the number of molecules of $NO\text{ and }O_2$

In mixture 1 : There are 5 $NO$ and 4 $O_2$ molecules.

In mixture 2 : There are 7 $NO$ and 2 $O_2$ molecules.

In mixture 3 : There are 3 $NO$ and 5 $O_2$ molecules.

Now we have to determine the rate law expression for mixture 1, 2 and 3.

The rate law expression for mixture 1 is:

$Rate=k[NO]^2[O_2]$

$Rate=k(5)^2\times (4)$

$Rate=k(100)$

The rate law expression for mixture 2 is:

$Rate=k[NO]^2[O_2]$

$Rate=k(7)^2\times (2)$

$Rate=k(98)$

The rate law expression for mixture 3 is:

$Rate=k[NO]^2[O_2]$

$Rate=k(3)^2\times (5)$

$Rate=k(45)$

Hence, the mixture 1 has the fastest initial rate.

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3 years ago

Calculate the number of moles found in 3.045x1024 atoms of helium.<br><br> PLS HELP

Kisachek [45]

Explanation:

so for this u have to use this equation where

Moles = number of particle/6.02×10^23

= 3.045 × 10^24/6.02×10^23

= 5.0581

write it to 3 S.F so 5.06 moles

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3 years ago

What is the relationship between DNA and proteins?

gayaneshka [121]

Explanation:

At first sight, I could recognize you

As if we were calling for each other

The DNA in my blood vessels tell me

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3 years ago

He rate constant of a reaction is 4.55 × 10−5 l/mol·s at 195°c and 8.75 × 10−3 l/mol·s at 258°c. what is the activation energy o

Xelga [282]

Answer : The activation energy of the reaction is, $17.285\times 10^4kJ/mole$

Solution :

The relation between the rate constant the activation energy is,

$\log \frac{K_2}{K_1}=\frac{Ea}{2.303\times R}\times [\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = initial rate constant = $4.55\times 10^{-5}L/mole\text{ s}$

$K_2$ = final rate constant = $8.75\times 10^{-3}L/mole\text{ s}$

$T_1$ = initial temperature = $195^oC=273+195=468K$

$T_2$ = final temperature = $258^oC=273+258=531K$

R = gas constant = 8.314 kJ/moleK

Ea = activation energy

Now put all the given values in the above formula, we get the activation energy.

$\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}]$

$Ea=17.285\times 10^4kJ/mole$

Therefore, the activation energy of the reaction is, $17.285\times 10^4kJ/mole$

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3 years ago

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What mass, in grams, of oxygen gas (o2) is contained in a 8.5-liter tank at 32.0 degrees celsius and 3.23 atmospheres?

e-lub [12.9K]

For this problem we assume that oxygen is an ideal gas. So, we use the equation PV=nRT where P is pressure, V is the volume, n is the number of moles, R is a universal constant and T is the temperature. We first solve for the number of moles n. Then, using the molar mass of oxygen we convert it to grams.

PV=nRT
n = PV / RT
n = 3.23 (8.5) / 0.08206 (32+273.15)
n = 1.0964 mol

mass = 1.0964 mol (32g / 1 mol) = 35.09 g O2

5 0

3 years ago