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WARRIOR [948]
4 years ago
15

Suppose you have a system made up of water only, with the container and everything beyond being the surroundings. Consider a pro

cess in which the water is first evaporated and then condensed back into its original container.
Is this two-step process necessarily reversible?
Chemistry
2 answers:
Airida [17]4 years ago
8 0

Answer:

Yes

Explanation:

The possibility of evaporating and condensing is a proof of reversible reaction

natima [27]4 years ago
6 0

Answer:

No

Explanation:

No, because there is no evidence that the surroundings are also restored to their original state.

cannot be undone by exactly reversing the change to the system; Spontaneous processes are irreversible.

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Ivenika [448]

Answer:

its always attractive

Explanation:

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5 0
2 years ago
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Consider the following reaction between mercury(II) chloride and oxalate ion.
Alina [70]

<u>Answer:</u> The rate law of the reaction is \text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

2 HgCl_2(aq.)+C_2O_4^{2-}(aq.)\rightarrow 2Cl^-(aq.)+2CO_2(g)+Hg_2Cl_2(s)

Rate law expression for the reaction:

\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b

where,

a = order with respect to HgCl_2

b = order with respect to C_2O_4^{2-}

Expression for rate law for first observation:

3.2\times 10^{-5}=k(0.164)^a(0.15)^b  ....(1)

Expression for rate law for second observation:

2.9\times 10^{-4}=k(0.164)^a(0.45)^b  ....(2)

Expression for rate law for third observation:

1.4\times 10^{-4}=k(0.082)^a(0.45)^b  ....(3)

Expression for rate law for fourth observation:

4.8\times 10^{-5}=k(0.246)^a(0.15)^b  ....(4)  

Dividing 2 from 1, we get:

\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{(0.164)^a(0.45)^b}{(0.164)^a(0.15)^b}\\\\9=3^b\\b=2

Dividing 2 from 3, we get:

\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{(0.164)^a(0.45)^b}{(0.082)^a(0.45)^b}\\\\2=2^a\\a=1

Thus, the rate law becomes:

\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2

3 0
3 years ago
Consider the substances hydrogen (H2), fluorine (F2), and hydrogen fluoride (HF). Based on their molecular structures, how does
katovenus [111]

Answer:

The boiling point of HF is <u><em>higher than</em></u> the boiling point of H2, and it is <u><em>higher than</em></u> the boiling point of F2.

Explanation:

In HF, inter- molecule forces will be present between the hydrogen and fluorine atoms. There will be hydrogen bonding present among the hydrogen and fluorine atoms. Hydrogen bonds are strong bonds and hence the boiling point for HF would be high as much energy will be required to break these bonds.

H2 and F2 will only have intra-molecular attractions and there will be no hydrogen bonds present in them. As a result, their boiling point will be lower.

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Number of grams in 4 mols of cu(no2)2
Drupady [299]
151.55 x 4 you have to add the molar masses for each element together.
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How cows are contributing to climate change.​
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they are apart of the life cycle

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