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just olya [345]
3 years ago
6

Determine the kinetic energy of a 150 kg roller coaster car that is moving with a speed of 15 m/s

Chemistry
1 answer:
morpeh [17]3 years ago
8 0

Answer:

K.E = 19125 j

Explanation:

Given data:

Mass = 150 Kg

Speed of car = 15 m/s

K.E = ?

Solution:

Formula:

K.E = 1/2 (mv²)

K.E = Kinetic energy

m = given mass

V = speed

now we will put the values in formula.

K.E = 1/2 (mv²)

K.E = 1/2 (150 Kg) (15²)

K.E = 1/2 150 Kg × 225 m²/s²

K.E = 0.5 × 225 m²/s² ×150 Kg

K.E = 19125 j

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Calculate the enthalpy of the reaction 4B(s)+3O2(g)→2B2O3(s) given the following pertinent information: B2O3(s)+3H2O(g)→3O2(g)+B
Ivenika [448]

Answer:

-2546 kJ

Explanation:

It is possible to obtain the enthalpy of a reaction from the sum of different intermediate reactions.

For the reaction:

4B(s) + 3O₂(g) → 2B₂O₃(s)

The intermediate reactions are:

A- B₂O₃(s) + 3H₂O(g) → 3O₂(g) + B₂H₆(g), ΔH°A= +2035 kJ

B- 2B(s) + 3H₂(g) → B₂H₆(g), ΔH°B= +36 kJ

C- H₂(g) + 1/2O₂(g) → H₂O(l), ΔH°C= -285 kJ

D- H₂O(l) → H₂O(g), ΔH°D= +44 kJ

2B = 4B(s) + 6H₂(g) → 2B₂H₆(g) ΔH°2B= +78 kJ

-2A = 6O₂(g) + 2B₂H₆(g) → 2B₂O₃(s) + 6H₂O(g) ΔH°-2A= -4070 kJ

-6C = 6H₂O(l) → 6H₂(g) + 3O₂(g) ΔH°-6C= +1710 kJ

-6D = 6H₂O(g) → 6H₂O(l) ΔH°-6D = -264 kJ

The sum of 2B - 2A - 6C - 6D produce:

4B(s) + 3O₂(g) → 2B₂O₃(s)

And the enthalpy is: ΔH°2B + ΔH°-2A + ΔH°-6C + ΔH°-6D = <em>-2546 kJ</em>

I hope it helps!

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