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LUCKY_DIMON [66]
2 years ago
13

Which statement is true about the relationship between entropy and spontaneity? O Spontaneous reactions tend to lead to higher e

ntropy. O Entropy is the same as spontaneity. As entropy increases, spontaneity decreases. o As entropy increases, spontaneity is unaffected. nics?
Chemistry
1 answer:
dimulka [17.4K]2 years ago
3 0

Answer:

The true statement is: Spontaneous reactions tend to lead to higher entropy.

Explanation:

The spontaneity of a reaction is linked to the value of Gibbs free energy (ΔG°). The more negative is this value, the more spontaneous is a reaction. At the same time, Gibbs free energy depends on enthalpy (ΔH°) and entropy (ΔS°), according to the following expression:

ΔG° = ΔH° - T.ΔS°

We can see that higher entropies (higher ΔS°) lead to more negative ΔG°, thus, more spontaneous reactions.

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What element/compounds are used in the melting of ice?
Galina-37 [17]

The temperature of the air, pavement, and the type of ice-melt compound used will affect the rate at which the ice melts. There are many different ice-melt compounds available from traditional rock salt (sodium chloride) to ice-melt pellets (calcium chloride).

8 0
3 years ago
How many valence electrons are in an atom of magnesium?
dimulka [17.4K]
There are two valence electrons in a single atom of magnesium.
8 0
3 years ago
Read 2 more answers
The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way:
LiRa [457]

Explanation:

(a)  As the given chemical reaction equation is as follows.

           OCl^{-} + I^{-} \rightarrow OI^{-1} + Cl^{-1}

So, when we double the amount of hypochlorite or iodine then the rate of the reaction will also get double. And, this reaction is "first order" with respect to hypochlorite and iodine.

Hence, equation for rate law of reaction will be as follows.

              Rate = K \times [OCl^{-}] \times [l^{-}]

(b)  Since, the rate equation is as follows.

                    Rate = K [OCl^{-}][l^{-}]

Let us assume that ([OCl^{-}] = [l^{-}])

Putting the given values into the above equation as follows.

             1.36 \times 10^{-4} = K \times (1.5 \times 10^{-3})^2

            1.36 \times 10^{-4} = K \times (2.25 \times 10^{-6})

                   K = \frac{1.36 \times 10^{-4}}{2.25 \times 10^{-6}}

                      = 60.4 M^{-1}sec^{-1}

Hence, the value of rate constant for the given reaction is 60.4 M^{-1}sec^{-1} .

(c) Now, we will calculate the rate as follows.

                Rate = K [OCl^{-}][l^{-}]

                         = 60.4 \times (1.8 \times 10^{3}) \times (6.0 \times 10^{4})

                        = 6.52 \times 10^{5}

Therefore, rate when [OCl^{-}] = 1.8 \times 10^{3} M and [I^{-}]= 6.0 \times 10^{4} M is  6.52 \times 10^{5}.

8 0
2 years ago
How many moles of hydrogen ions are formed in the ionization of 0.250 moles of H2SO4?
True [87]

Answer:

The ionization of 0.250 moles of H₂SO₄ will produce 0.5 moles of H⁺ (hydrogen ion)

Explanation:

From the ionization of H₂SO₄, we have

H₂SO₄ → 2H⁺ + SO₄²⁻

Hence, at 100% yield, one mole of H₂SO₄ produces two moles of H⁺ (hydrogen ion) and one mole of SO₄²⁻ (sulphate ion), therefore, 0.250 moles of H₂SO₄ will produce 2×0.250 moles of H⁺ (hydrogen ion) or 0.5 moles of H⁺ (hydrogen ion) and 0.25 moles of SO₄²⁻ (sulphate ion).

That is; 0.250·H₂SO₄ → 0.5·H⁺ + 0.250·SO₄²⁻.

4 0
3 years ago
How many grams of O₂ are required to react completely with 14.6 g of Na to form sodium oxide, Na₂O?
Bad White [126]

The balanced chemical reaction is :

O_2 + 4Na \ -> \ 2Na_2O

Number of moles of Na, n = \dfrac{14.6}{23} = 0.635 \  mol .

Now, from balance chemical reaction we can see that 1 mole of oxygen reacts with 4 moles of sodium.

So, number of moles of oxygen are :

n = \dfrac{0.635}{4}\  mole

So, amount of oxygen required is :

m = \dfrac{0.635 \times 32}{4}\  gm\\\\m = 5.08 \ gm

Therefore, 5.08 gram of oxygen will react with 14.6 gram of sodium.

7 0
2 years ago
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