If an automobile air bag has a volume of 11.7 L, what mass of NaN3 (in g) is required to fully inflate the air bag upon impact

Chemistry

1 answer:

Usimov [2.4K]3 years ago

5 0

Answer:

33.95 grams of NaN3

Explanation:

Number of moles of NaN3 = mass (m)/MW = m/65 mole

I mole of NaN3 requires 22.4L air bag

m/65 mole of NaN3 required 11.7L

22.4m/65 = 11.7

22.4m = 65×11.7

22.4m = 760.5

m = 760.5/22.4 = 33.95grams of NaN3

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3 years ago

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3 years ago

Alveoli are tiny sacs of air in the lungs. Their average diameter is 4.50 × 10−5 m. Calculate the uncertainty in the velocity of

Arisa [49]

Answer: The uncertainty in the velocity of oxygen molecule is $4.424\times 10^{-5}m/s$

Explanation:

The diameter of the molecule will be equal to the uncertainty in position.

The equation representing Heisenberg's uncertainty principle follows:

$\Delta x.\Delta p=\frac{h}{2\pi}$

where,

$\Delta x$ = uncertainty in position = d = $4.50\times 10^{-5}m$

$\Delta p$ = uncertainty in momentum = $m\Delta v$

m = mass of oxygen molecule = $5.30\times 10^{-26}kg$

h = Planck's constant = $6.627\times 10^{-34}kgm^2/s^2$

Putting values in above equation, we get:

$4.50\times 10^{-5}m\times 5.30\times 10^{-26}kg\times \Delta v=\frac{6.627\times 10^{-34}kgm^2/s}{2\times 3.14}\\\\\Delta v=\frac{6.627\times 10^{-34}kgm^2/s^2}{2\times 3.14\times 4.50\times 10^{-5}m\times 5.30\times 10^{-26}kg}=4.424\times 10^{-5}m/s$