All categories

3 years ago

Use the information below to explain why the atomic radius decreases across a period.

Chemistry

1 answer:

pishuonlain [190]3 years ago

7 0

Answer:

Detail is given below.

Explanation:

Atomic radius trend along period:

As we move from left to right across the periodic table the number of valance electrons increases. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required.

we can see in picture A there is one positive charge and force of attraction is 2.30×10⁻⁸ N

In picture D there are 2 positive charge and force of attraction is 4.60×10⁻⁸ N

So by increasing the positive charge force of attraction also increases.

In F we can see there is more positive charge thus force of attraction is also greater.

You might be interested in

When 0.0901 mol of an unknown hydrocarbon is burned in a bomb calorimeter, the calorimeter increases in temperature by 2.19°C. I

Alexus [3.1K]

Answer:

The heat of combustion for the unknown hydrocarbon is -29.87 kJ/mol

Explanation:

Heat capacity of the bomb calorimeter = C = 1.229 kJ/°C

Change in temperature of the bomb calorimeter = ΔT = 2.19°C

Heat absorbed by bomb calorimeter = Q

$Q=C\times \Delta T$

$Q=1.229 kJ/^oC\times 2.19^oC=2,692 kJ$

Moles of hydrocarbon burned in calorimeter = 0.0901 mol

Heat released on combustion = Q' = -Q = -2,692 kJ

The heat of combustion for the unknown hydrocarbon :

$\frac{Q'}{0.090 mol}=\frac{-2,692 kJ}{0.0901 mol}=-29.87 kJ/mol$

6 0

3 years ago

Read 2 more answers

A chemist prepares a solution of sodium thiosulfate (Na2S203) by measuring out 110.g of sodium thiosulfate into a 350. mL volume

forsale [732]

Answer:

1.99 M

Explanation:

The molar mass of sodium thiosulfate (solute) is 158.11 g/mol. The moles corresponding to 110 grams are:

110 g × (1 mol/158.11 g) = 0.696 mol

The volume of solution is 350 mL = 0.350 L.

The molarity of sodium thiosulfate is:

M = moles of solute / liters of solution

M = 0.696 mol / 0.350 L

M = 1.99 M

5 0

3 years ago

if 14.0 g of aluminium reacts with excess sulfuric acid to produce 75.26 g of aluminium sulfate, what is the percent yield?

vlada-n [284]

Taking into account definition of percent yield, the percent yield for the reaction is 84.88%.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 Al + 3 H₂SO₄ → Al₂(SO₄)₃ + 3 H₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Al: 2 moles
H₂SO₄: 3 moles
Al₂(SO₄)₃. 1 mole
H₂: 3 moles

The molar mass of the compounds is:

Al: 27 g/mole
H₂SO₄: 98 g/mole
Al₂(SO₄)₃: 342 g/mole
H₂: 2 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Al: 2 moles ×27 g/mole= 54 grams
H₂SO₄: 3 moles ×98 g/mole= 294 grams
Al₂(SO₄)₃: 1 mole ×342 g/mole= 342 grams
H₂: 3 moles ×2 g/mole= 6 grams

<h3>Mass of aluminium sulfate formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 54 grams of aluminium form 342 grams of aluminium sulfate, 14 grams of aluminium form how much mass of aluminium sulfate?

$mass of aluminium sulfate=\frac{14 grams of aluminium x342 grams of aluminium sulfate}{54 grams of aluminium}$

<u><em>mass of aluminium sulfate= 88.67 grams</em></u>

Then, 88.67 grams of aluminium sulfate can be produced if 14.0 g of aluminium reacts with excess sulfuric acid.

<h3>Percent yield</h3>

The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.

The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

$percent yield=\frac{actual yield}{theorical yield} x100$

where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

<h3>Percent yield for the reaction in this case</h3>

In this case, you know: