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3 years ago

What is the difference between the ground state and excited state for an atom

Chemistry

1 answer:

Juliette [100K]3 years ago

7 0

Answer:

The ground state describes the lowest possible energy that an atom can have.

Explanation:

An excited state is an energy level of an atom, ion, or molecule in which an electron is at a higher energy level than its ground state.

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rate of a certain reaction is given by the following rate law: rate Use this information to answer the questions below. What is

Sunny_sXe [5.5K]

Complete Question

The rate of a certain reaction is given by the following rate law:

$rate = k [H_2][I_2]$

rate Use this information to answer the questions below.

What is the reaction order in $H_2$ ?

What is the reaction order in $I_2$ ?

What is overall reaction order?

At a certain concentration of H2 and I2, the initial rate of reaction is 2.0 x 104 M / s. What would the initial rate of the reaction be if the concentration of H2 were doubled? Round your answer to significant digits. The rate of the reaction is measured to be 52.0 M / s when [H2] = 1.8 M and [I2] = 0.82 M. Calculate the value of the rate constant. Round your answer to significant digits.

Answer:

The reaction order in $H_2$ is n = 1

The reaction order in $I_2$ is m = 1

The overall reaction order z = 2

When the hydrogen is double the the initial rate is $rate_n = 4.0*10^{-4} M/s$

The rate constant is $k = 35.23 \ M^{-1} s^{-1}$

Explanation:

From the question we are told that

The rate law is $rate = k [H_2][I_2]$

The rate of reaction is $rate = 2.0 *10^{4} M /s$

Let the reaction order for $H_2$ be n and for $I_2$ be m

From the given rate law the concentration of $H_2$ is raised to the power of 1 and this is same with $I_2$ so their reaction order is n=m=1

The overall reaction order is

$z = n +m$

$z =1 +1$

$z =2$

At $rate = 2.0 *10^{4} M /s$

$2.0*10^{4} = k [H_2] [I_2] ---(1)$

= > $k = \frac{2.0*10^{4}}{[H_2] [I_2] }$

given that the concentration of hydrogen is doubled we have that

$rate = k [2H_2] [I_2] ----(2)$

=> $k = \frac{rate_n }{ [2H_2] [I_2]}$

So equating the two k

$\frac{2.0*10^{4}}{[H_2] [I_2] } = \frac{rate_n }{ [2H_2] [I_2]}$

=> $rate_n = 4.0*10^{-4} M/s$

So when

$rate_x = 52.0 M/s$

$[H_2] = 1.8 M$

$[I_2] = 0.82 \ M$

We have

$52 .0 = k(1.8)* (0.82)$

$k = \frac{52 .0}{(1.8)* (0.82)}$

$k = 35.23 M^{-2} s^{-1}$

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2 years ago

3. Balance each of the following redox reactions in the listed aqueous environment. (4 pts each, 8 pts total) Crs)NO3 (a) Cr (a)

GrogVix [38]

Explanation:

(a) The given reaction equation is as follows.

$Cr(s) + NO^{-}_{3}(aq) \rightarrow Cr^{3+}(aq) + NO(g)$ (acidic)

So, here the reduction and oxidation-half reactions will be as follows.

Oxidation-half reaction: $Cr(s) \rightarrow Cr^{3+}(aq) + 3e^{-}$

Reduction-half-reaction: $NO^{-}_{3} + 3e^{-}(aq) \rightarrow NO(g)$

As total charge present on reactant side is -1 and total charge present on product side is +3. And, since it is present in aqueous medium. Hence, we will balance the charge for this reaction equation as follows.

$Cr(s) + NO^{-}_{3}(aq) + 4H^{+}(aq) \rightarrow Cr^{3+}(aq) + NO(g) + 2H_{2}O(l)$ (acidic)

(b) The given reaction equation is as follows.

$HCO^{-}_{3}(aq) + Ag(s) + NH_{3}(aq) \rightarrow H_{2}CO(aq) + Ag(NH_{3})^{+}_{2}(aq)$ (basic)

So, here the reduction and oxidation-half reactions will be as follows.

Reduction-half reaction: $HCO^{-}_{3}(aq) + 4e^{-} \rightarrow H_{2}CO(aq)$

Oxidation-half reaction: $Ag(s) \rightarrow Ag(NH_{3})^{+}_{2}(aq) + 1e^{-}$

Hence, to balance the number of electrons in this equation we multiply it by 4 as follows.

$4Ag(s) \rightarrow 4Ag(NH_{3})^{+}_{2}(aq) + 4e^{-}$

Therefore, balancing the whole reaction equation in the basic medium as follows.

$H_{2}CO(aq) + 4Ag(NH_{3})^{+}_{2}(aq) + 5OH^{-}(aq) \rightarrow HCO^{-}_{3}(aq) + 4Ag(s) + 8NH_{3}(aq) + 3H_{2}O(l)$

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3 years ago