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3 years ago

Rank the nonmetals in each set from most reactive (1) to least reactive (3). Neon: Selenium: Fluorine

Chemistry

2 answers:

Ganezh [65]3 years ago

4 0

Neon:

Selenium:

Fluorine:

izzie

2 years ago

That is correct :)

expeople1 [14]3 years ago

3 0

The periodic trend of reactivity depends on whether the elements are metals or nonmetals. For metals, reactivity increases down a group and from right to left across a period. Nonmetals do the opposite. For nonmetals, reactivity increases up a group and from left to right across a period. Therefore, the rank is as follows:

Fluorine

Selenium

Neon

Hope this answers the question. Have a nice day.

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Helga [31]

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Explanation:

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2 years ago

how many grams of calcium sulfate are produced from 10 grams of calcium nitrate and how many grams of calcium sulfate are produc

AlekseyPX

Answer: 8.30 g of calcium sulfate are produced from 10 grams of lithium sulfate.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} Ca(NO_3)_2=\frac{10g}{164g/mol}=0.061moles$

$\text{Moles of} Li_2SO_4=\frac{10g}{110g/mol}=0.091moles$

$Ca(NO_3)_2+Li_2SO_4\rightarrow 2LiNO_3+CaSO_4$

According to stoichiometry :

1 mole of $Ca(NO_3)_2$ require = 1 mole of $Li_2SO_4$

Thus 0.061 moles of $Ca(NO_3)_2$ will require= $\frac{1}{1}\times 0.061=0.061moles$ of $Li_2SO_4$

Thus $Ca(NO_3)_2$ is the limiting reagent as it limits the formation of product and $Li_2SO_4$ is the excess reagent.

As 1 mole of $Ca(NO_3)_2$ give = 1 mole of $CaSO_4$

Thus 0.061 moles of $Ca(NO_3)_2$ give = $\frac{1}{1}\times 0.061=0.061moles$ of $CaSO_4$

Mass of $CaSO_4=moles\times {\text {Molar mass}}=0.061moles\times 136g/mol=8.30g$

Thus 8.30 g of calcium sulfate are produced from 10 grams of lithium sulfate.

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3 years ago

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AnnyKZ [126]

The answer would actually be false. I just took the test.

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3 years ago

Read 2 more answers

What is the mole fraction of methanol in a solution that contains 6.0 mol of methanol and 3.0 mol of water? The formula for meth

Ivenika [448]

The mole fraction of a product is the number of moles of the product divided by the total number of moles of the solution.

Here moles of methanol = 6.0 moles

Moles of solution = 6.0 moles of methanol + 3.0 moles of water = 9.0 moles of solution

Mole fraction of methanol = 6.0 / 9.0 = 0.67

Answer: 0.67

4 0

3 years ago

Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refi

Fed [463]

Answer:

A. -166.6 kJ/mol

B. -127.7 kJ/mol

C. -133.9 kJ/mol

Explanation:

Let's consider the oxidation of sulfur dioxide.

2 SO₂(g) + O₂(g) → 2 SO₃(g) ΔG° = -141.8 kJ

The Gibbs free energy (ΔG) can be calculated using the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature (25 + 273.15 = 298.15 K)

Q is the reaction quotient

The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:

$[]=\frac{P}{R.T}$

Calculate ΔG at 25°C given the following sets of partial pressures.

Part A 130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M$

$[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln (4.44 × 10⁻⁵) = -166.6 kJ/mol

Part B 5.0atm SO₂, 3.0atm O₂, 30atm SO₃ Express your answer using four significant figures.

$[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M$

$[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M$

$[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol

Part C Each reactant and product at a partial pressure of 1.0 atm. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 24.4 = -133.9 kJ/mol

7 0

3 years ago