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Masteriza [31]
2 years ago
13

5.1 mol is equal to_____particles.

Chemistry
1 answer:
Amanda [17]2 years ago
8 0

Answer:

3.07122*10^24 particles

Explanation:

To convert moles to particles, remember that a mole of something is defined as 6.022*10^23 atoms of that thing.

5.1 mol * 6.022x10^23/1 mol = 3.07122*10^24 particles

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Entropy increase from solid to liquid to gas. T/F
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Answer: True

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

\Delta S is positive when randomness increases and \Delta S is negative when randomness decreases.

When the solid changes to liquid to gas ,the randomness of particles increases and hence \Delta S is positive and thus the entropy increases.

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Which of the following is a cause of eutrophication of a lake?
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In one sentence, describe the general chemical structure of carbohydrates. In a second sentence, list the major biochemical func
Brums [2.3K]

Answer:

A carbohydrate is a compound composed of molecules of carbon (C), hydrogen (H), and oxygen (O) such that its general chemical formula is CH_2OH.

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Explanation:

A carbohydrate is a compound composed of molecules of carbon (C), hydrogen (H), and oxygen (O) such that its general chemical formula is CH_2OH.

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6 0
2 years ago
Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g)
Fynjy0 [20]

Answer:

Rate of formation of SO₃ [\frac{d[SO_{3}] }{dt}] = 7.28 x 10⁻³ M/s

Explanation:

According to equation   2 SO₂(g) + O₂(g) → 2 SO₃(g)

Rate of disappearance of reactants = rate of appearance of products

                     ⇒ -\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}  -----------------------------(1)

    Given that the rate of disappearance of oxygen = -\frac{d[O_{2} ]}{dt} = 3.64 x 10⁻³ M/s

             So the rate of formation of SO₃ [\frac{d[SO_{3}] }{dt}] = ?

from equation (1) we can write

                                   \frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]

                                ⇒ \frac{d[SO_{3}] }{dt} = 2 x 3.64 x 10⁻³ M/s

                                ⇒ [\frac{d[SO_{3}] }{dt}] = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ [\frac{d[SO_{3}] }{dt}] = 7.28 x 10⁻³ M/s

7 0
3 years ago
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