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4 years ago

Draw and label the structure of a sodium atom showing the distribution of electrons

Chemistry

1 answer:

Korolek [52]4 years ago

8 0

Answer:

Please, find the drawing in the picture attached.

Explanation:

A simple model or drawing of the structure of an atom may be performed showing the electrons on the different energy levels (shells), plus the nucleons (protons and neutrons).

Then, you need the information on:

number of protons (atomic number)
electron configuration
number of neutrons

The first step is to find the atomic number of the atom. You find that information in the periodic table, where the elements are show with their chemical symbol.

The chemical symbol of sodium is Na. It is located in the group (colum) 1 and period (raw) 3. There you find that the atomic number is 11. Hence your drawing must show 11 protons.

Since the atom is neutral, the number of electrons is also 11. So, you must place the electrons in the following energy levels or shells:

Main energy level shell number of electrons

n = 1 K 2 (this is the maximum in this level)

n= 2 L 8 (this is the maximun in this level)

n= 3 M 1 (the maximum in this level is 18)

Now that you have the number of protons and the location of the electrons, you must determine the number of neutrons.

In the periodict table you will also find the atomic mass of the elements. The atomic mass of sodium is 22.990 amu. That means that most sodium atoms have mass number (number of protons plus neutrons) 23. Some isotopes will have less neutrons, but you can draw your model using the most typical isotope with 23 nucleons. This means, that the number of neutrons is 23 nucleons - 11 protons = 12 neutrons.

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The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: OCl−+I−→OI−+Cl− T

motikmotik

Answer :

(a) The rate law for the reaction is:

$\text{Rate}=k[OCl^-]^1[I^-]^1$

(b) The value of rate constant is, $60.4M^{-1}s^{-1}$

(c) rate of the reaction is $6.52\times 10^{-5}Ms^{-1}$

Explanation :

Rate law : It 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:

$OCl^-+I^-\rightarrow OI^-+Cl^-$

Rate law expression for the reaction:

$\text{Rate}=k[OCl^-]^a[I^-]^b$

where,

a = order with respect to $OCl^-$

b = order with respect to $I^-$

Expression for rate law for first observation:

$1.36\times 10^{-4}=k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b$ ....(1)

Expression for rate law for second observation:

$2.72\times 10^{-4}=k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b$ ....(2)

Expression for rate law for third observation:

$2.72\times 10^{-4}=k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b$ ....(3)

Dividing 1 from 2, we get:

$\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}\\\\2=2^a\\a=1$

Dividing 1 from 3, we get:

$\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b}\\\\2=2^b\\b=1$

Thus, the rate law becomes:

$\text{Rate}=k[OCl^-]^a[I^-]^b$

a = 1 and b = 1

$\text{Rate}=k[OCl^-]^1[I^-]^1$

Now, calculating the value of 'k' (rate constant) by using any expression.

$1.36\times 10^{-4}=k(1.5\times 10^{-3})(1.5\times 10^{-3})$

$k=60.4M^{-1}s^{-1}$

Now we have to calculate the rate for a reaction when concentration of $OCl^-$ and $I^-$ is $1.8\times 10^{-3}M$ and $6.0\times 10^{-4}M$ respectively.