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Vika [28.1K]
2 years ago
5

As a rock falls, some of its potential energy is transformed into kinetic energy. What is true about the mechanical energy durin

g this transformation?
A.
It decreases because potential energy is a kind of mechanical energy.
B.
It does not change because kinetic energy and potential energy are both kinds of mechanical energy.
C.
It increases because kinetic energy is a kind of mechanical energy.
D.
It does not change because kinetic energy and potential energy are not kinds of mechanical energy.
Chemistry
1 answer:
mars1129 [50]2 years ago
4 0

Answer:

a or b

Explanation:

because those are the most likely to be true

now it gives you two options so i thinks it might be B but you could choose A you got two options choose wisely

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The conversion of cyclopropane to propene occurs with a first-order rate constant of . How long will it take for the concentrati
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This is an incomplete question, here is a complete question.

The conversion of cyclopropane to propene occurs with a first-order rate constant of 2.42 × 10⁻² hr⁻¹. How long will it take for the concentration of cyclopropane to decrease from an initial concentration 0.080 mol/L to 0.053 mol/L?

Answer : The time taken will be, 17.0 hr

Explanation :

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 2.42\times 10^{-2}\text{ hr}^{-1}

t = time passed by the sample  = ?

a = initial concentration of the reactant  = 0.080 M

a - x = concentration left = 0.053 M

Now put all the given values in above equation, we get

t=\frac{2.303}{2.42\times 10^{-2}}\log\frac{0.080}{0.053}

t=17.0\text{ hr}

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Answer:

131.5 kJ

Explanation:

Let's consider the following reaction.

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First, we will calculate the standard enthalpy of the reaction (ΔH°).

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g) ) - 1 mol × ΔH°f(CaCO₃(s) )

ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)

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Then, we calculate the standard entropy of the reaction (ΔS°).

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g) ) - 1 mol × S°(CaCO₃(s) )

ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)

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Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.

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