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valentina_108 [34]
2 years ago
15

What kind of solid often has the highest melting points?

Chemistry
2 answers:
ki77a [65]2 years ago
7 0

Solids made of atoms

Explanation: Apex

vovikov84 [41]2 years ago
6 0

Answer:

ones with ionic bonds

Explanation:

they have high melting and boiling points

You might be interested in
CHEMISTRY HELP PLEASE *answer all questions please*
Lerok [7]

Answer:

<u><em>Question 1: </em></u>

A) 0.289 moles.

B) 1.74 x 10²³ atoms.

<u><em>Question 2:</em></u>

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.

<u><em>Question 3:</em></u>

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:

<u><em>Question 1:</em></u>

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

<em>using cross multiplication: </em>

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.


<u><em>Question 2:</em></u>

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.

<em><u>using cross multiplication: </u></em>

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

<em><u>using cross multiplication:</u></em>

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.


<u><em>Question 3:</em></u>

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.

  • <em><u>using cross multiplication: </u></em>

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

<em><u>using cross multiplication: </u></em>

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
2 years ago
Acid base reaction is called nuetralization reaction why?​
Lesechka [4]

Answer:

Explanation:

You have an acid that is acidic or a base that is basic. When you mix the two, they form water (assuming those are bronsted-lowry acids and bases) which is neutral.

3 0
3 years ago
Read 2 more answers
How many moles are in 9.12 x 1023 molecules of sugar?​
aleksandrvk [35]

Answer:

1.514 moles

Explanation:

For this problem you want to use dimensional analysis and cancel out your molecules of sugar and be left with moles of sugar. We know that 1 mole (of anything) = 6.022 x 10 ^ 23 molecules, so we should use that conversion to help us. Start with 9.12 x 10 ^23 molecules and divide by 6.022 x 10 ^ 23 molecules, and you will be left with moles.

Hope this helps!

3 0
2 years ago
If you had excess chlorine, how many moles of of aluminum chloride could be produced from 19.0 g of aluminum?
SIZIF [17.4K]
The chemical reaction would be written as follows:

2Al + 3Cl2 = 2AlCl3

We are given the amount of aluminum to be used in the reaction. This will be the starting point of the calculations. We do as follows:

19.0 g Al ( 1 mol / 29.98 g ) ( 2 mol AlCl3 / 2 mol Al ) = 0.63 mol AlCl3
5 0
3 years ago
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Akimi4 [234]

Answer:

d I think there's two answer

Explanation:

or c

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