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3 years ago

Atom with the largest atomic radius in group 18

Chemistry

1 answer:

Mkey [24]3 years ago

8 0

Answer:

Radon

Explanation:

Group eighteen elements are called noble gases. This group includes helium, neon, argon, krypton, xenon and radon.

Radon is present at the last of this column.

Atomic radii trend along group:

As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.

As the size of atom increases the ionization energy from top to bottom also decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus.

Thus radon has largest atomic radius.

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How many moles are in 9.25E24 formulas units of sodium acetate?

Lelu [443]

Answer:

<h2>15.37 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{9.25 \times {10}^{24} }{6.02 \times {10}^{23} } \\ = 15.365448...$

We have the final answer as

<h3>15.37 moles</h3>

Hope this helps you

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3 years ago

This is the chemical formula for Chormium (III) nitrate Cr(NO3)3Calculate the mass percent of nitrogen in chromium(III) nitrate.

butalik [34]

Answer:

The percent composition for nitrogen in chromium(III) nitrate is 6%.

Explanation:

6 0

3 years ago

Dilute 0.407M & 2.56L solution to 7.005L. CF?

Alenkinab [10]

Answer:

C₂ = 0.149 M

Explanation:

Given data:

Initial concentration = 0.407 M

Initial volume = 2.56 L

Final volume = 7.005 L

Final concentration = ?

Solution:

Formula:

C₁V₁ = C₂V₂

C₁ = Initial concentration

V₁ = Initial volume

C₂ = Final concentration

V₂ =Final volume

Now we will put the values.

0.407 M × 2.56 L = C₂ × 7.005 L

1.042 = C₂ × 7.005 L

C₂ = 1.042 M.L / 7.005 L

C₂ = 0.149 M

5 0

3 years ago

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Vsevolod [243]

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3 0

3 years ago

Read 2 more answers

Balance the equation and show the calculation of the number of moles and grams of CO2 formed from 11.9 grams of O2. Show your an

n200080 [17]

Answer:

0.2349 moles, 10.3356 g

Explanation:

The given reaction is :

$C_6H_{14}+O_2\rightarrow CO_2 + H_2O$

The balanced reaction by equating the same number of each atom both side is :

$2C_6H_{14}+19O_2\rightarrow 12CO_2 + 14H_2O$

Given that :

Amount of oxygen gas = 11.9 g

Molar mass of oxygen gas = 32 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus, moles are:

$moles= \frac{11.9\ g}{32\ g/mol}$

$moles= 0.3719\ mol$

From the reaction,

19 moles of oxygen gas on reaction forms 12 moles of carbon dioxide

Also,

1 mole of oxygen gas on reaction forms 12/19 moles of carbon dioxide

So,

0.3719 moles of zinc on reaction forms $\frac {12}{19}\times 0.3719$ mole of carbon dioxide

<u>Moles of carbon dioxide formed = 0.2349 moles</u>

Mass of carbon dioxide = moles×Molar mass

Molar mass of carbon dioxide = 44 g/mol

<u>Mass of carbon dioxide formed = 0.2349 ×44 g = 10.3356 g</u>

4 0

3 years ago