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Amanda [17]
3 years ago
13

Which of the halide ions (f−, cl−, br−, and i−) is the most stable base? which is the least stable base?

Chemistry
2 answers:
rusak2 [61]3 years ago
6 0

Explanation:

It is known that on moving down a group there will occur a decrease in electronegativity of non-metals. Whereas stability of their conjugate bases increases on moving down the group.

This also means that their acid strength also increases.

For example, HX + H_{2}O \rightarrow H_{3}O^{+} + X^{-}

where, X = F, Cl, Br or I)

Therefore, acidity will increase in the following order.

                      HI > HBr > HCl > HF

Hence, the stability of their conjugate bases will be as follows.

                I^{-} > Br^{-} > Cl^{-} > F^{-}

Thus, we can conclude that the most stable base is I^{-} and least stable base is F^{-}.

Nezavi [6.7K]3 years ago
3 0
You should take note that the question is about stability. A compound is stable if it does not easily react with other elements. Hence, its reactivity must be low. As you move down the group, reactivity decreases. So, the halide at the very bottom is the least reactive. It would then be logical that the most stable conjugate base is I⁻ and the least stable conjugate base is the most reactive which is F⁻.
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Crystal field splitting is the difference in energy between d orbitals of ligands. Crystal field splitting number is denoted by the capital Greek letter Δ. Crystal field splitting explains the difference in color between two similar metal-ligand complexes.

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3 0
2 years ago
NaCl+AgC2H3O2--&gt; NaC2H3O2+AgCl <br><br><br> what is this balanced
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Answer:

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3 0
2 years ago
For the reaction 2N2O5(g) &lt;---&gt; 4NO2(g) + O2(g), the following data were colected:
KonstantinChe [14]

Answer:

a) The reaction is first order, that is, order 1. Option C is correct.

b) The half life of the reaction is 23 minutes. Option B is correct

c) The initial rate of production of NO2 for this reaction is approximately = (3.7 × 10⁻⁴) M/min. Option has been cut off.

Explanation:

First of, we try to obtain the order of the reaction from the data provided.

t (minutes) [N2O5] (mol/L)

0 1.24x10-2

10 0.92x10-2

20 0.68x10-2

30 0.50x10-2

40 0.37x10-2

50 0.28x10-2

70 0.15x10-2

Using a trial and error mode, we try to obtain the order of the reaction. But let's define some terms.

C₀ = Initial concentration of the reactant

C = concentration of the reactant at any time.

k = rate constant

t = time since the reaction started

T(1/2) = half life

We Start from the first guess of zero order.

For a zero order reaction, the general equation is

C₀ - C = kt

k = (C₀ - C)/t

If the reaction is indeed a zero order reaction, the value of k we will obtain will be the same all through the set of data provided.

C₀ = 0.0124 M

At t = 10 minutes, C = 0.0092 M

k = (0.0124 - 0.0092)/10 = 0.00032 M/min

At t = 20 minutes, C = 0.0068 M

k = (0.0124 - 0.0068)/20 = 0.00028 M/min

At t = 30 minutes, C = 0.0050 M

k = (0.0124 - 0.005)/30 = 0.00024 M/min

It's evident the value of k isn't the same for the first 3 trials, hence, the reaction isn't a zero order reaction.

We try first order next, for first order reaction

In (C₀/C) = kt

k = [In (C₀/C)]/t

C₀ = 0.0124 M

At t = 10 minutes, C = 0.0092 M

k = [In (0.0124/0.0092)]/10 = 0.0298 /min

At t = 20 minutes, C = 0.0068 M

k = 0.030 /min

At t = 30 minutes, C = 0.0050 M

k = 0.0303

At t = 40 minutes

k = 0.0302 /min

At t = 50 minutes,

k = 0.0298 /min

At t = 60 minutes,

k = 0.031 /min

This shows that the reaction is indeed first order because all the answers obtained hover around the same value.

The rate constant to be taken will be the average of them all.

Average k = 0.0302 /min.

b) The half life of a first order reaction is related to the rate constant through this relation

T(1/2) = (In 2)/k

T(1/2) = (In 2)/0.0302

T(1/2) = 22.95 minutes = 23 minutes.

c) The initial rate of production of the product at the start of the reaction

Rate = kC (first order)

At the start of the reaction C = C₀ = 0.0124M and k = 0.0302 /min

Rate = 0.0302 × 0.0124 = 0.000374 M/min = (3.74 × 10⁻⁴) M/min

3 0
2 years ago
3) During the day at 27°C a cylinder with a sliding top contains 20.0 liters
tatiyna

Answer:

T_2=12\°C

Explanation:  

Hello there!  

In this case, according to the Charles' law equation which help us to understand the directly proportional relationship between volume and temperature:

\frac{T_2}{V_2}=\frac{T_1}{V_1}  

Thus, by solving for the final temperature, T2, and making sure we use the temperatures in Kelvin, we can calculate the final temperature as shown below:

T_2=\frac{T_1V_2}{V_1}  \\\\T_2=\frac{(27+273)K*19L}{20.0L}\\\\T_2=285-273\\\\T_2=12\°C

Best regards!  

Best regards!

4 0
2 years ago
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