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andreyandreev [35.5K]
3 years ago
9

how does an airplane’s kinetic energy and potential energy change as it takes off and lands? How does this energy change relate

to the law of conservation of energy?
Chemistry
1 answer:
mojhsa [17]3 years ago
8 0

Answer: As the airplane goes higher, the mechanical energy is changed into gravitational potential energy. While flying, some energy is lost through drag to thermal (heat) energy and sound energy. Some is also lost as the plane makes the air around it move. ... As speed and height decrease, kinetic and potential energy decrease.

Explanation:

Hope this help

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Bohr looked to improve upon Rutherford’s model of the atom because Bohr thought that Rutherford’s model:
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Write a chemical equation for nh4+(aq) showing how it is an acid or a base according to the arrhenius definition.
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An Arrhenius acid by definition dissociates in water to form H3O+ (or H+) ions while an arrhenius base dissociates in water to form OH- ions.

NH4+(aq) can be categorised as an arrhenius acid since it releases H3O+ ions in aqueous media

NH4+(aq) + H2O (aq) ↔ NH3 (aq) + H3O+(aq)

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Nitrogen gas reacts with hydrogen gas to produce ammonia. How many liters of hydrogen gas at 95kPa and 15∘C are required to prod
viktelen [127]

Answer:

222.30 L

Explanation:

We'll begin by calculating the number of mole in 100 g of ammonia (NH₃). This can be obtained as follow:

Mass of NH₃ = 100 g

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= 14 + 3

= 17 g/mol

Mole of NH₃ =?

Mole = mass /molar mass

Mole of NH₃ = 100 / 17

Mole of NH₃ = 5.88 moles

Next, we shall determine the number of mole of Hydrogen needed to produce 5.88 moles of NH₃. This can be obtained as follow:

N₂ + 3H₂ —> 2NH₃

From the balanced equation above,

3 moles of H₂ reacted to produce 2 moles NH₃.

Therefore, Xmol of H₂ is required to p 5.88 moles of NH₃ i.e

Xmol of H₂ = (3 × 5.88)/2

Xmol of H₂ = 8.82 moles

Finally, we shall determine the volume (in litre) of Hydrogen needed to produce 100 g (i.e 5.88 moles) of NH₃. This can be obtained as follow:

Pressure (P) = 95 KPa

Temperature (T) = 15 °C = 15 + 273 = 288 K

Number of mole of H₂ (n) = 8.82 moles

Gas constant (R) = 8.314 KPa.L/Kmol

Volume (V) =?

PV = nRT

95 × V = 8.82 × 8.314 × 288

95 × V = 21118.89024

Divide both side by 95

V = 21118.89024 / 95

V = 222.30 L

Thus the volume of Hydrogen needed for the reaction is 222.30 L

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Lelechka [254]
I believe Positive & Negative
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Answer:

A is the correct option

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