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Hatshy [7]
2 years ago
8

If 5.32 mols N2 and 15.8 mols H2 react together, what mass NH3 can be

Chemistry
1 answer:
hodyreva [135]2 years ago
3 0

Answer:

2.87 gram

N2 is the limiting agent

Explanation:

We will find out if there is sufficient N2 and h2 to produce NH3

a) For 2.36 grams of N2

Molar mass of N2 = 28.02

Number of moles of N2 in 2.36 grams = 2.36/28.02

Mass of NH3 = 17.034 g

Now NH3 produced form 2.36 grams of N2 =  

2.36/28.02 * 2 * 17.034 = 2.87 g NH3

b) For 1.52 g of H2  

NH3 produced = 1.52/2.016 * (2/3) * 17.034 = 8.56

N2 Is not enough to produce 2.87 g of NH3 and also H2 is not enough to make 8.56 g of NH3.  

N2 is the limiting agent as it has smaller product mass

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The Only Viable Solution: A New Moon Shot

The electric light bulb didn't appear from efforts to develop better candles or telling people to use less light.

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g Consider an ideal atomic gas in a cylinder. The upper part of the cylinder is a moveable piston of negligible weight. The heig
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A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

A cylindrical weight with a mass (m) of 3 kg is dropped, that is, its initial velocity (u) is 0 m/s and travels 10 m (s). Assuming the acceleration (a) is that of gravity (9.8 m/s²). We can calculate the velocity (v) of the weight in the instant prior to the collision with the piston using the following kinematic equation.

v^{2} = u^{2} + 2as = 2 (9.8 m/s^{2} ) (10m) \\\\v = 14 m/s

The object with a mass of 3 kg collides with the piston at 14 m/s, The kinetic energy (K) of the object at that moment is:

K = \frac{1}{2} m v^{2} = \frac{1}{2} (3kg) (14m/s)^{2} = 294 J

The kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Thus, Q = 294 J.

Given all the process is at 250 K (T), we can calculate the change of entropy of the gas using the following expression.

\Delta S_{gas} = \frac{Q}{T} = \frac{294 J}{250K} = 1.18 J/K

The change in the entropy of the environment, has the same value but opposite sign than the change in the entropy of the gas. Thus, \Delta S_{env} = -1.18 J/K

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

Learn more: brainly.com/question/22655760

6 0
2 years ago
A sample of helium gas occupies a volume of 152.0 mL at a pressure of 717.0 mm Hg and a temperature of 315.0 K. What will the vo
kozerog [31]

Answer:

0.581 L  or  581 mL

Explanation:

As stated in the question, the combined gas law is (P1*V1/T1) = (P2*V2/T2)

Write down the amounts you are given.

V1 = 0.152 L (I was taught to always convert milliliters to liters)

P1 = 717 mmHg

T1 = 315 K

V2 = ?

P2 = 463 mmHg

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(717 mmHg*0.152 L) / (315 K) = (463 mmHg*V2) / (777 K)

0.346 = (463*V2) / (777)

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Pressure and volume are indirectly proportional. This checks out because the volume increased while pressure decreased. Volume and temperature are directly proportional. This checks out because both volume and temperature increased. This is a good way to check your answers. You can also solve each side of the combined gas law equation to see if they are both the same.

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How many grams are in 0.35 moles of C2H4? please show dimensional analysis
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Answer:

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Using molar mass times moles of the chemical to find the mass in 0.35 moles of c2h4:

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Answer:

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