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spin [16.1K]
3 years ago
8

One way to remove the pollutant nitrogen monoxide, NO, from mobile diesel exhaust is by reacting it with ammonia, NH, as seen in

the balanced reaction below. What mass, in grams, of NH, is required to remove 57.0 grams of NO from the air? 4 NH3 + 6 NO → 5 N, + 6H2O​
Chemistry
1 answer:
SSSSS [86.1K]3 years ago
6 0

Answer:

m_{NH_3}=21.6gNH_3

Explanation:

Hello there!

In this case, in agreement to the given chemical reaction, it is possible for us to calculate the mass of NH3 required to remove 57.0 g NO via the stoichiometry based off the 4:6 mole ratio between them:

m_{NH_3}=57.0g NO*\frac{1molNO}{30.01gNO}*\frac{4molNH_3}{6molNO}  *\frac{17.03gNH_3}{1molNH_3} \\\\m_{NH_3}=21.6gNH_3

Best regards!

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The density of liquid mercury is 13.5 g/cm^3. What mass of mercury (in kg) is required to fill a hollow cylinder having an inner
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Answer: 1.06 kg

Explanation:

Density is defined as the mass contained per unit volume.

Density=\frac{mass}{Volume}

Given :

Density of mercury = 13.5g/cm^3

volume of cylinder  = volume of mercury = \pir^2h

where r = radius of cylinder = \frac{diameter}{2}=\frac{2cm}{2}=1cm

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Putting in the values we get:

volume of mercury =3.14\times 1^2\times 25=78.5cm^3

mass=density\times volume=13.5g/cm^3\times 78.5cm^3=1059.75g=1.06kg      (1kg=1000g)

Thus mass of mercury required to fill a hollow cylinder having an inner diameter of 2.00 cm to a height of 25.0 cm is 1.06 kg.

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