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Marianna [84]
3 years ago
6

The air bags in cars are inflated when a collision triggers the explosive, highly exothermic decomposition of sodium azide (NaN3

): 2NaN3(s) → 2Na(s) + 3N2(g) The passenger-side air bag in a typical car must fill a space approximately four times as large as the driver-side bag to be effective. Calculate the mass of sodium azide required to fill a 113-L air bag. Assume the pressure in the car is 1.00 atm and the temperature of N2 produced is 85°C.
Chemistry
1 answer:
Oksanka [162]3 years ago
3 0

Answer : The mass of NaN_3 required is, 166.4 grams.

Explanation :

First we have to calculate the moles of nitrogen gas.

Using ideal gas equation:

PV=nRT

where,

P = Pressure of N_2 gas = 1.00 atm

V = Volume of N_2 gas = 113 L

n = number of moles N_2 = ?

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

T = Temperature of N_2 gas = 85^oC=273+85=358K

Putting values in above equation, we get:

1.00atm\times 113L=n\times (0.0821L.atm/mol.K)\times 358K

n=3.84mol

Now we have to calculate the moles of sodium azide.

The balanced chemical reaction is,

2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)

From the balanced reaction we conclude that

As, 3 mole of N_2 produced from 2 mole of NaN_3

So, 3.84 moles of N_2 produced from \frac{2}{3}\times 3.84=2.56 moles of NaN_3

Now we have to calculate the mass of NaN_3

\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3

Molar mass of NaN_3 = 65 g/mole

\text{ Mass of }NaN_3=(2.56moles)\times (65g/mole)=166.4g

Therefore, the mass of NaN_3 required is, 166.4 grams.

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Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down, and it has a height of 20 cm
8090 [49]

Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

\frac{dV}{dt}=10\frac{cm^3}{s}

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

\frac{dh}{dt} = ?*\frac{dV}{dt}

Of course, ?=\frac{dh}{dV}.

Now, since the volume of a cone is V=\pi r^2h/3 and the ratio r/h=15/20=3/4 or r=3/4h, the volume becomes:

V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3

We proceed to its differentiation:

\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}

Then, we compute \frac{dh}{dt}

\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}

Finally, at h=2:

\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}

Best regards.

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