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3 years ago

How many grams of nitrogen gas are needed to produce 34 g of ammonia

Physics

1 answer:

scZoUnD [109]3 years ago

7 0

Answer:

28 grams

Explanation:

The equation for the reaction is

3H(2) + N(2) -> 2NH(3)

Then we have.

The molar mass, M of ammonia is 17 g/mol.

34 grams of ammonia, NH3 then would be

34 g / 17 g/mol

= 2 moles

2 moles of ammonia will be obtained from

(2 * 1) / 2

= 1 mole of nitrogen

The molar masses of nitrogen is 28 g/mol

2 moles of nitrogen corresponds to 1 * 28 = 28 grams.

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Electrical generator

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2 years ago

Which part of the electromagnetic spectrum is divided into seven ranges of wavelengths and is the part visible to the human eye

oksano4ka [1.4K]

In one of the most amazing coincidences in all of science,
the part of the electromagnetic spectrum that's visible to the
human eye is called "visible light".

Visible light is not 'divided' into anything. We mention the names
to seven of the colors in visible light. But all of the thousands of
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3 years ago

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Control rods in a nuclear reactor are used to:

kiruha [24]

B.absorb neutrons to prevent chain reactions which become uncontrollable

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2 years ago

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An object of mass 3.4 kg is moving in a straight line with kinetic energy 59.177 J. A force is applied in the direction of its m

rusak2 [61]

Answer:

Its momentum is multiplied by a factor of 1.25

Explanation:

First, we calculate the initial velocity of the object:

K = 0.5 * m * v₁²

59.177 J = 0.5 * 3.4 kg * v₁²

v₁ = 5.9 m/s

With that velocity we can calculate the initial momentum of the object:

p₁ = v₁ * m

p₁ = 20.06 kg·m/s

Then we calculate the velocity of the object once its kinetic energy has increased:

(59.177 J) * 1.57 = 0.5 * 3.4 kg * v₂²

v₂ = 7.4 m/s

And calculate the second momentum of the object:

p₂ = v₂ * m

p₂ = 25.16 kg·m/s

Finally we calculate the factor:

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3 0

2 years ago

Estimate the length of the neptunian year using the fact that the earth is 1.50Ã108km from the sun on average.

Ksenya-84 [330]

Answer:

Explanation:

We know that,

Neptune is 4.5×10^9 km from the sun

And given that,

Earth is 1.5×10^8km from sun

Then,

Let P be the orbital period and

Let a be the semi-major axis

Using Keplers third law

Then, the relation between the orbital period and the semi major axis is

P² ∝ a³

Then,

P² = ka³

P²/a³ = k

So,

P(earth)²/a(earth)³ = P(neptune)² / a(neptune)³

Period of earth P(earth) =1year

Semi major axis of earth is

a(earth) = 1.5×10^8km

The semi major axis of Neptune is

a (Neptune) = 4.5×10^9km

So,

P(E)²/a(E)³ = P(N)² / a(N)³

1² / (1.5×10^8)³ = P(N)² / (4.5×10^9)³

Cross multiply

P(N)² = (4.5×10^9)³ / (1.5×10^8)³

P(N)² = 27000

P(N) =√27000

P(N) = 164.32years

The period of Neptune is 164.32years

6 0

3 years ago