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miv72 [106K]
3 years ago
14

Automobile air bags use the decomposition of sodium azide as their sources of gas for rapid inflation, represented in the reacti

on below. What mass (in grams) of NaN3 is required to provide 40.0 L of N2 at 25°C and 763 torr?
Physics
1 answer:
NeTakaya3 years ago
5 0

Answer : The mass of NaN_3 required is 71.175 grams.

Explanation :

To calculate the moles of nitrogen gas, we use the equation given by ideal gas :

PV = nRT

where,

P = Pressure of nitrogen gas = 763 torr

V = Volume of the nitrogen gas = 40.0 L

n = number of moles of gas = ?

R = Gas constant = 62.364\text{ L.torr }mol^{-1}K^{-1}

T = Temperature of helium gas = 25^oC=273+25=298K

Putting values in above equation, we get:

763torr\times 40.0L=n\times 62.364\text{ L.torr }mol^{-1}K^{-1}\times 298K\\\\n=1.642mol

Now we have to calculate the moles of NaN_3.

The balanced chemical reaction is:

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

From the balanced chemical reaction, we conclude that

As, 3 moles of N_2 produced from 2 moles NaN_3

So, 1.642 moles of N_2 produced from \frac{2}{3}\times 1.642=1.095 moles NaN_3

Now we have to calculate the mass of NaN_3.

Molar mass of NaN_3 = 65 g/mol

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

\text{Mass of }NaN_3=1.095mole\times 65g/mole=71.175g

Therefore, the mass of NaN_3 required is 71.175 grams.

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2 years ago
Water with a volume flow rate of 0.001 m3/s, flows inside a horizontal pipe with diameter of 1.2 m. If the pipe length is 10m an
galben [10]

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\triangle P=1.95*10^{-4}

Explanation:

Mass m=0.001

Diameter d=1.2m

Length l=10m

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 Q=AV

 V=\frac{Q}{\pi/4D^2}

 V=\frac{0.001}{\pi/4(1.2)^2}

 V=8.84*10^{-4}

Generally the equation for Friction factor is mathematically given by

 F=\frac{64}{Re}

Where Re

Re=Reynolds Number

 Re=\frac{pVD}{\mu}

 Re=\frac{1000*8.84*10^{-4}*1.2}{1.002*10^{-3}}

 Re=1040

Therefore

 F=\frac{64}{Re}

 F=\frac{64}{1040}

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Generally the equation for Friction factor is mathematically given by

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 H=\frac{0.06*10*(8.9*10^-4)^2}{2*1.2*9.81}

 H=19.9*10^{-9}

Where

H=\frac{\triangle P}{\rho g}

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6 0
3 years ago
A concave lens has a focal length of 20 cm. A real object is 30 cm from the lens. Where is the image? What is the magnification?
Pani-rosa [81]

Answer:

12 cm and 0.4

Explanation:

f = - 20 cm, u = - 30 cm

Let v be the position of image and m be the magnification.

Use lens equation

1 / f = 1 / v - 1 / u

- 1 / 20 = 1 / v + 1 / 30

1 / v = - 5 / 60

v = - 12 cm

m = v / u = - 12 / (-30) = 0.4

6 0
3 years ago
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