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3 years ago

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Which 10-milliliter sample of water has the greatest degree of disorder?A)H2O(g) at 120°CB)H2O() at 80°CC)H2O() at 20°CD)H2O(s)

at 0°C

1 answer:

AnnyKZ [126]3 years ago

6 0

Answer:

A) H₂O at 120°C

Explanation:

It is possible to think the higher temperature, the greatest degree of disorder. That is because with a high temperature, vibrations of molecules increases.

In general, at low temperatures, the molecules are in solid state (The lowest degree of disorder), increasing its temperature, molecules becomes in liquids, and, with more temperature, are gases (The greatest degree of disorder).

Thus, the sample that has the greatest degree of disorder is:

<h3>A) H₂O at 120°C</h3>

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Be sure to answer all parts. Express the rate of reaction in terms of the change in concentration of each of the reactants and p

Tanzania [10]

Answer : The [H] is increasing at the rate of 0.36 mol/L.s

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$2D(g)+3E(g)+F(g)\rightarrow 2G(g)+H(g)$

The expression for rate of reaction :

$\text{Rate of disappearance of }D=-\frac{1}{2}\frac{d[D]}{dt}$

$\text{Rate of disappearance of }E=-\frac{1}{3}\frac{d[E]}{dt}$

$\text{Rate of disappearance of }F=-\frac{d[F]}{dt}$

$\text{Rate of formation of }G=+\frac{1}{2}\frac{d[G]}{dt}$

$\text{Rate of formation of }H=+\frac{d[H]}{dt}$

$\text{Rate of reaction}=-\frac{1}{2}\frac{d[D]}{dt}=-\frac{1}{3}\frac{d[E]}{dt}=-\frac{d[F]}{dt}=+\frac{1}{2}\frac{d[G]}{dt}=+\frac{d[H]}{dt}$

Given:

$-\frac{d[D]}{dt}=0.18mol/L.s$

As,

$-\frac{1}{2}\frac{d[D]}{dt}=+\frac{d[H]}{dt}=0.18mol/L.s$

and,

$+\frac{d[H]}{dt}=2\times 0.18mol/L.s$

$+\frac{d[H]}{dt}=0.36mol/L.s$

Thus, the [H] is increasing at the rate of 0.36 mol/L.s

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Why is a bowling ball harder to lift than a basket ball? Is it bc a bowling ball is made of metal?

vagabundo [1.1K]

It is more slippery, and it is heavier

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Consider the first-order reaction described by the equation At a certain temperature, the rate constant for this reaction is 5.8

zubka84 [21]

<u>Answer:</u> The half life of the reaction is 1190.7 seconds

<u>Explanation:</u>

The equation used to calculate rate constant from given half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

where,

k = rate constant of the reaction = $5.82\times 10^{-4}s^{-1}$

$t_{1/2}$ = half life of the reaction = ?

Putting values in above equation, we get:

$t_{1/2}=\frac{0.693}{5.82\times 10^{-4}s^{-1}}=1190.7s$

Hence, the half life of the reaction is 1190.7 seconds

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How many moles of NH3 can be produced from 30.0 mol of H2 and excess N2?

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weqwewe [10]

Answer:

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