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2 years ago

The boiling point of chlorine is −34 °C. Which of the following best predicts the boiling point of iodine?

Chemistry

2 answers:

Luda [366]2 years ago

5 0

Answer:

Higher than −34 °C because dispersion forces are stronger in iodine due to a greater number of electrons.

Explanation:

Setler79 [48]2 years ago

4 0

Answer:

a) Higher than −34 °C because dispersion forces are stronger in iodine due to a greater number of electrons.

Explanation:

Both chlorine and Iodine are halogens. They belong to the 7th group on the periodic table.

To examine the trend of boiling point in this group we must consider the nature of the intermolecular bonds that would be formed by the molecules of these elements. We know that London dispersion forces, a type of van der Waals attraction would be more prevalent. This is so because the molecules here would be non-polar and the uneven distribution of the constantly moving electrons would initiate the intermolecular bonding.

The uneven charge distribution leads to the formation of a temporary dipole.

This makes boiling point increase down the group because more electrons becomes involved and the bond becomes stronger.

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Methane and ethane are both made up of carbon and hydrogen. In methane, there are 12.0 g of carbon for every 4.00 g of hydrogen,

Sati [7]

Answer:

The answer is: Law of multiple proportions

Explanation:

The law of multiple proportions is a law of chemical combination given by Dalton in 1803.

According to this law, if more than one chemical compound is formed by combining two elements, then the mass of an element that combines with the fixed mass of other element is represented in the form of small whole number ratio.

Therefore, is an illustration of the law of the law of multiple proportions.

8 0

2 years ago

The complex ion, [ni(nh3)6] 2+, has a maximum absorption near 580 nm. calculate the crystal field splitting energy (in kj/mol) f

Wewaii [24]

In given data:
maximum absorption wavelength λ = 580 nm = 580 x 10⁻⁹ m
write the equation to find the crystal field splitting energy:
E = hC / λ
Here, E is the crystal field splitting energy, h = 6.63 x 10⁻³⁴ J.sec is Planck's constant and C = 3 x 10⁸ m/sec is speed of light.
substitute in the equation above:
E = (6.64 x 10⁻³⁴ x 3 x 10⁸) / (580 x 10⁻⁹) = 3.43 x 10⁻¹⁹J
This crystal field splitting energy is for 1 ion.
Number of atoms in one mole, NA = 6.023 x 10²³
to calculate the crystal field splitting energy for one mole:
E(total) = E x NA
= (3.43 x 10⁻¹⁹) x (6.023 x 10²³) = 206 kJ/ mole

5 0

2 years ago

Write down the dissolution equation for rubidium chromate dissolving in water. (Chromate is a polyatomic ion with the formula Cr

ycow [4]

Answer:

Four moles of the cation

Explanation:

2Rb2CrO4(s)<--------> 4Rb^+(aq) + 2CrO4^2-(aq)

Now looking at the reaction equation, it can be seen that one mole of rubidium chromate contains two moles of rubidium ions and one mole of chromate ions.

The dissolution of two moles of rubidium chromate should then yield four moles of rubidium ions and two moles of chromate ions since the ratio of ions present is 2:1.

This explains the reaction equation written above for the dissolution of two moles of rubidium chromate as shown.

5 0

2 years ago

10.0 mL of a 0.100 mol L–1 solution of a metal ion M2+ is mixed with 10.0 mL of a 0.100 mol L–1 solution of a substance L. The f

Mrac [35]

The chemical reaction is
 M2+(aq) + 2L(aq) <==> ML22+(aq)
Intial concentration 0.10 0.10
Change -x -2x +x
Equilibrium 0.10 - x 0.01 = 0.10 - 2x x
Solving for x
0.01 = 0.10 - 2x
x = 0.045
The equilibrium concentration of ML22+ is 0.045 mol L-1

7 0

2 years ago

Read 2 more answers

A student reacts 20.0 mL of 0.248 M H2SO4 with 15.0 mL of 0.195 M NaOH. Write a balanced chemical equation to show this reaction

Molodets [167]

Answer: The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For NaOH:

Initial molarity of NaOH solution = 0.195 M

Volume of solution = 15.0 mL = 0.015 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.195M=\frac{\text{Moles of NaOH}}{0.015L}\\\\\text{Moles of NaOH}=(0.195mol/L\times 0.015L)=2.925\times 10^{-3}mol$

For sulfuric acid:

Initial molarity of sulfuric acid solution = 0.248 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.248M=\frac{\text{Moles of }H_2SO_4}{0.020L}\\\\\text{Moles of }H_2SO_4=(0.248mol/L\times 0.020L)=4.96\times 10^{-3}mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, $2.925\times 10^{-3}$ moles of KOH will react with = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sulfuric acid

As, given amount of sulfuric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, NaOH is considered as a limiting reagent because it limits the formation of product.

Excess moles of sulfuric acid = $(4.96-1.462)\times 10^{-3}=3.498\times 10^{-3}mol$

By Stoichiometry of the reaction:

2 moles of KOH produces 1 mole of sodium sulfate

So, $2.925\times 10^{-3}$ moles of KOH will produce = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sodium sulfate

For sodium sulfate:

Moles of sodium sulfate = $1.462\times 10^{-3}moles$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sodium sulfate}=\frac{1.462\times 10^{-3}}{0.035}=0.0418M$

For sulfuric acid:

Moles of excess sulfuric acid = $3.498\times 10^{-3}mol$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sulfuric acid}=\frac{3.498\times 10^{-3}}{0.035}=0.0999M$

For NaOH:

Moles of NaOH remained = 0 moles

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of NaOH}=\frac{0}{0.050}=0M$

Hence, the concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

8 0

2 years ago