What is characteristic of a covalent bond?

A binary compound of boron and hydrogen has the following percentage composition: 78.14% boron, 21.86% hydrogen. If the molar ma

algol [13]

Answer:

Empirical formula: BH3

Molecular Formula: B2H6

Explanation:

To solve the exercise, we need to know how many boron atoms and how many hydrogen atoms the compound has. We know that of the total weight of the compound, 78.14% correspond to boron and 21.86% to hydrogen. As the weight of the compound is between 27 g and 28 g, using the above percentages we can solve that the compound has between 21.1 g and 21.8 g of boron, and between 5.9 g and 6.1 g of hydrogen:

100% _____ 27 g

78.14% _____ x = 78.14% * 27g / 100% = 21.1 g boron

100% ______27 g

21.86% ______ x = 21.86% * 27g / 100% = 5.9 g hydrogen

100% _____ 28 g

78.14% _____ x = 78.14% * 28g / 100% = 21.8 g boron

100% _____ 28g

21.86% _____ x = 21.86% * 28g / 100% = 6.1 g hydrogen

So, if the atomic weight of boron is 10.8 g, there must be two boron atoms in the compound that sum 21.6 g. The weight of hydrogen is 1 g, so the compound must have six hydrogen atoms.

The molecular formula represents the real amount of atoms that form a compound. Therefore, the molecular formula of the compound is B2H6.

The empirical formula is the minimum expression that represents the proportion of atoms in a compound. For example, ethane has 2 carbon atoms and 6 hydrogen atoms, so its molecular formula is C2H6, however, its empirical formula is CH3. Therefore, the empirical formula of the boron compound is BH3.

8 0

3 years ago

CHEMISTRY HELP PLEASE *answer all questions please*

Lerok [7]

Answer:

Question 1:

A) 0.289 moles.

B) 1.74 x 10²³ atoms.

Question 2:

A) 0.30 moles.

B) it contains 0.3 moles of both Na and Cl.

C) it contains 6.023 x 10²³ atoms of both Na and Cl.

Question 3:

A) The number of moles of sucrose (C₁₂H₂₂O₁₁) ≅ 0.0228 moles.

B) The number of moles of C atoms in sucrose (C₁₂H₂₂O₁₁) = 0.2763 mole of C atoms.

The number of moles of H atoms in sucrose (C₁₂H₂₂O₁₁) = 0.5016 mole of H atoms.

The number of moles of O atoms in sucrose (C₁₂H₂₂O₁₁) = 0.2508 mole of O atoms.

C) The number of C atoms = 1.65 x 10²³ atoms.

The number of H atoms = 3.02 x 10²³ atoms.

The number of O atoms = 1.51 x 10²³ atoms.

Explanation:

Question 1:

A) The number of moles of Au in 57.01 g sample:

n = mass / molar mass,

mass = 57.01 g and molar mass = 196.966 g/mol e.

The number of moles of Au in the sample = (57.01 g) / (196.966 g/mole) = 0.289 moles.

B) The number of atoms of Au in the sample:

It is known that every mole of a substance contains Avogadro,s number (NA = 6.023 x 10²³) of molecules.

1.0 mole of Au → 6.023 x 10²³ atoms

0.289 mole of Au → ???? atoms

using cross multiplication:

The number of atoms of Au in the sample = (6.023 x 10²³ x 0.289 mole) / (1.0 mole) = 1.74 x 10²³ atoms.

Question 2:

A) The number of moles of 17.45 g of NaCl:

n = mass / molar mass,

mass = 17.45 g and molar mass = 58.44 g/mole.

The number of moles of NaCl = (17.45 g) / (58.44 g/mole) = 0.298 mole ≅ 0.30 moles.

B) The number of moles of each element in NaCl

NaCl → Na + Cl

Each mole of NaCl contains one mole of Na and one mole of Cl.

using cross multiplication:

1.0 mole NaCl → 1.0 mole Na

0.3 mole NaCl → ??? mole Na

The number of moles of Na atoms in NaCl = (1.0 mole Na x 0.3 mole NaCl) / (1.0 mole NaCl) = 0.3 mole of Na atoms.

by the same way; the number of moles of Cl atoms = (1.0 mole Cl x 0.3 mole NaCl) / (1.0 mole NaCl) = 0.3 mole of Cl atoms.

C) The number of atoms of each element in the sample:

It is known that every mole of a substance contains Avogadro,s number (NA = 6.023 x 10²³) of molecules.

1.0 mole of NaCl → 6.023 x 10²³ molecules

0.3 mole of NaCl → ???? molecules

using cross multiplication:

The number of molecules in 0.3 mole of NaCl = (6.023 x 10²³ x 0.3 mole) / (1.0 mole) = 1.8069 x 10²³ molecules.

Every molecule of NaCl contains one atom of Na and one atom of Cl.

So, it contains 6.023 x 10²³ atoms of both Na and Cl.

Question 3:

A) The number of moles of 7.801 g of sucrose (C₁₂H₂₂O₁₁):

n = mass / molar mass,

mass = 7.801 g and molar mass = 342.3 g/mole.

The number of moles of sucrose (C₁₂H₂₂O₁₁) = (7.801 g) / (342.3 g/mol) = 0.022789 mol ≅ 0.0228 moles.

B) The number of moles of each element in sucrose (C₁₂H₂₂O₁₁):

C₁₂H₂₂O₁₁ → 12C + 22H + 11O

Each mole of sucrose contains 12 moles of C, 22 moles of H, and 11 moles of O.

using cross multiplication:

1.0 mole of sucrose (C₁₂H₂₂O₁₁) → 12.0 moles C

0.0228 mole of sucrose (C₁₂H₂₂O₁₁) → ??? moles C

The number of moles of C atoms in sucrose (C₁₂H₂₂O₁₁) = (12.0 moles C x 0.0228 moles of sucrose (C₁₂H₂₂O₁₁)) / (1.0 mole sucrose (C₁₂H₂₂O₁₁)) = 0.2763 mole of C atoms.

By the same way; the number of moles of H atoms:

1.0 mole of sucrose (C₁₂H₂₂O₁₁) → 22.0 moles H

0.0228 mole of sucrose (C₁₂H₂₂O₁₁) → ??? moles H

The number of moles of H atoms in sucrose (C₁₂H₂₂O₁₁) = (22.0 moles H x 0.0228 moles of sucrose (C₁₂H₂₂O₁₁)) / (1.0 mole sucrose (C₁₂H₂₂O₁₁)) = 0.5016 mole of H atoms.

Also; the number of moles of O atoms:

1.0 mole of sucrose (C₁₂H₂₂O₁₁) → 11.0 moles O

0.0228 mole of sucrose (C₁₂H₂₂O₁₁) → ??? moles O

The number of moles of O atoms in sucrose (C₁₂H₂₂O₁₁) = (11.0 moles H x 0.0228 moles of sucrose (C₁₂H₂₂O₁₁)) / (1.0 mole sucrose (C₁₂H₂₂O₁₁)) = 0.2508 mole of O atoms.

C) The number of atoms of each element in the sucrose (C₁₂H₂₂O₁₁) sample:

It is known that every mole of a substance contains Avogadro,s number (NA = 6.023 x 10²³) of molecules.

1.0 mole of sucrose (C₁₂H₂₂O₁₁) → 6.023 x 10²³ molecules

0.0228 mole of sucrose (C₁₂H₂₂O₁₁) → ???? molecules

using cross multiplication:

The number of molecules in 0.0228 mole of sucrose (C₁₂H₂₂O₁₁) = (6.023 x 10²³ x 0.0228 mole) / (1.0 mole) = 1.273 x 10²² molecules.

Each molecule of sucrose contains 12 atoms of C, 22 atoms of H, and 11 atoms of O.

So, the number of each atom that the sucrose (C₁₂H₂₂O₁₁) sample contains are:

The number of C atoms = (12 x 1.273 x 10²² molecules) = 1.65 x 10²³ atoms.

The number of H atoms = (22 x 1.273 x 10²² molecules) = 3.02 x 10²³ atoms.

The number of O atoms = (11 x 1.273 x 10²² molecules) = 1.51 x 10²³ atoms.

6 0

3 years ago

Give an example that shows energy transfer and another one that shows energy transformation.

solong [7]

Answer:

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░░░▌░▄▄▄▐▌▀▀▀░░ This is Bob

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▀█▌░░░▄░▀█▀░▀ ░░ Copy And Paste Him onto all of ur brainly answers

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Explanation:

5 0

3 years ago

What is the number of atoms in 3.75 mol fe

Kruka [31]

1mol=6.022x10^23 atoms
3.75mol= 6.022x3.75x10^23 atoms
=2.2583x10^24 atoms

8 0

3 years ago

Is ok2 gas a element compound or homogeneous or heterogeneous mixture?

ivolga24 [154]

I'm guessing you mean O2 gas? This is an elemental compound. It can't be a mixture because it is composed of only 1 element. Hope I helped!

4 0

3 years ago