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3 years ago

what conclusions can be made about the relationship between metallic character and the atomic radius?

Chemistry

1 answer:

kolezko [41]3 years ago

5 0

We have to get the relationship between metallic character and atomic radius.

Metallic character increases with increase in atomic radius and decrease with decrease of atomic radius.

If electrons from outermost shell of an element can be removed easily, that atom can be considered to have more metallic character.

With increase in atomic radius, nuclear force of attraction towards outermost shell electron decreases which facilitates the release of electron.

With decrease in atomic radius, nuclear force of attraction towards outermost shell electrons increases, so electrons are hold tightly to nucleus. Hence, removal of electron from outermost shell becomes difficult making the atom less metallic in nature.

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What is the best description of cirrus clouds?

miskamm [114]

Option 1

Explanation:

Cirrus clouds are short, detached, hair-like clouds found at high altitudes

6 0

2 years ago

Read 2 more answers

Consider the following reaction: 4Fe + 302 → 2Fe2O3. What mass of iron(III) oxide would

frutty [35]

Answer:

Mass = 357.7 g

Explanation:

Given data:

Mass of Fe = 250 g

Mass of oxygen = 120 g

Mass of iron(III) oxide produced = ?

Solution:

Chemical equation:

4Fe + 3O₂ → 2Fe₂O₃

Number of moles of Fe:

Number of moles = mass/molar mass

Number of moles = 250 g/ 55.8 g/mol

Number of moles = 4.48 mol

Number of moles of O₂ :

Number of moles = mass/molar mass

Number of moles = 120 g/ 32 g/mol

Number of moles = 3.75 mol

Now we will compare the moles of reactants with product.

Fe : Fe₂O₃

4 : 2

4.48 : 2/4×4.48 = 2.24

O₂ : Fe₂O₃

3 : 2

3.75 : 2/3×3.75= 2.5

Less number of moles of Fe₂O₃ are produced by Fe thus it will act as limiting reactant.

Mass of Fe₂O₃:

Mass = number of moles × molar mass

Mass = 2.24 mol × 159.69 g/mol

Mass = 357.7 g

3 0

3 years ago

Plz help me with this question plzzzzz

LekaFEV [45]

Answer:

Explanation:

Formula: Al2O3

If we require 2Al2O3

We divide 2 by 3

Sorry for the shadow of phone and fingers.

8 0

3 years ago

How many liters of nitrogen gas are needed to make 25 mol of nitrogen trifluoride

ikadub [295]

The reaction between N₂ and F₂ gives Nitrogen trifluoride as the product. The balanced equation is;

N₂ + 3F₂ → 2NF₃

The stoichiometric ratio between N₂ and NF₃ is 1 : 2
Hence,
moles of N₂ / moles of F₂ = 1 / 2
moles of N₂ / 25 mol = 0.5
moles of N₂ = 0.5 x 25 mol = 12.5 mol

Hence N₂ moles needed = 12.5 mol

At STP (273 K and 1 atm) 1 mol of gas = 22.4 L

Hence needed N₂ volume = 22.4 L mol⁻¹ x 12.5 mol
= 280 L

6 0

3 years ago

An unknown gas Q requires 2.67 times as long to effuse under the same conditions as the same amount of nitrogen gas. What is the

Misha Larkins [42]

Answer:

The correct answer is 199.66 grams per mole.

Explanation:

Based on law of effusion given by Graham, a gas rate of effusion is contrariwise proportionate to the square root of molecular mass, that is, rate of effusion of gas is inversely proportional to the square root of mass. Therefore,

R1/R2 = √ M2/√ M1

Here rate is the rate of effusion of the gas expressed in terms of number of mole per uni time or volume, and M is the molecular mass of the gas.

Rate Q/Rate N2 = √M of N2/ √M of Q

The molecular mass of N2 or nitrogen gas is 28 grams per mole and M of Q is molecular mass of Q and based on the question Q needs 2.67 times more to effuse in comparison to nitrogen gas, therefore, rate of Q = rate of N2/2.67

Now putting the values we get,

rate of N2/2.67/rate of N2 = √28/ √M of Q

√M of Q = √ 28 × 2.67

M of Q = (√ 28 × 2.67)²

M of Q = 199.66 grams per mole

3 0

3 years ago