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Butoxors [25]
3 years ago
12

someone please explain how to separate copper powder from a mixture containing copper and zinc powder !!

Chemistry
1 answer:
jarptica [38.1K]3 years ago
6 0

Answer:

by burning the mixture in an electrical furnace to a temp. 640 degree C. to melt both Zinc & Aluminum leaving copper oxide as a powder.

Explanation:

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Given 4.80g of ammonium carbonate, find:
V125BC [204]

Answer:

1) 0.05 mol.

2) 0.1 mol.

3) 0.05 mol.

4) 0.4 mol.

5) 2.4 x 10²³ molecules.

Explanation:

<em>1) Number of moles of the compound:</em>

no. of moles of ammonium carbonate = mass/molar mass = (4.80 g)/(96.09 g/mol) = 0.05 mol.

<em>2) Number of moles of ammonium ions :</em>

  • Ammonium carbonate is dissociated according to the balanced equation:

<em>(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.</em>

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

<em>∴ The no. of moles of NH₄⁺ ions in 0.05 mol of (NH₄)₂CO₃ </em>= (2.0)(0.05 mol) =  <em>0.1 mol.</em>

<em>3) Number of moles of carbonate ions :</em>

  • Ammonium carbonate is dissociated according to the balanced equation:

<em>(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.</em>

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

∴ The no. of moles of CO₃²⁻ ions in 0.05 mol of (NH₄)₂CO₃ = (1.0)(0.05 mol) = 0.05 mol.

<em>4) Number of moles of hydrogen atoms:</em>

  • Every 1.0 mol of (NH₄)₂CO₃  contains:

2.0 moles of N atoms, 8.0 moles of H atoms, 1.0 mole of C atoms, and 3.0 moles of O atoms.

<em>∴ The no. of moles of H atoms in 0.05 mol of (NH₄)₂CO</em>₃ = (8.0)(0.05 mol) = <em>0.4 mol.</em>

<em>5) Number of hydrogen atoms:</em>

  • It is known that every mole of a molecule or element contains Avogadro's number (6.022 x 10²³) of molecules or atoms.

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

1.0 mole of H atoms contains → 6.022 x 10²³ atoms.

0.4 mole of H atoms contains → ??? atoms.

<em>∴ The no. of atoms in  0.4 mol of H atoms</em> = (6.022 x 10²³ molecules)(0.4 mole)/(1.0 mole) = <em>2.4 x 10²³ molecules.</em>

8 0
3 years ago
Write electron configurations for the following ion: Cd2 Cd2 . Express your answer in order of increasing orbital energy. For ex
Mashutka [201]

Answer:

Cd2+ : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10 or [Kr] 4d¹⁰

Explanation:

Before proceeding to write out the electron configuration of Cd2+, we have to obtain the electron configuration of Cadmium (Cd),

Cadmium has an atomic number of 48, this means that a neutral cadmium atom will have a total of  48  electrons surrounding its nucleus.

The electronic configuration of Cadmium is;

Cd: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10

The shorthand notation is given as;

Cd: [Kr] 4d¹⁰5s²

Cd2+ means that it has two less electrons, hence it's electron configuration is given as;

Cd2+ : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10 or [Kr] 4d¹⁰

5 0
3 years ago
3. 78 mL of 2.5 M phosphoric acid is neutralized with 500 mL of potassium hydroxide. What is the
xenn [34]

Hence, concentration of base is 1.17 M
7 0
3 years ago
The diagram below shows changes to the nucleus of an atom.
anzhelika [568]

Answer:

maybe 1 or 3 im not sure

Explanation:

i didn't study it yet sorry for not helping but try asking someone else

8 0
3 years ago
Read 2 more answers
Balance these scientific equations:
ivanzaharov [21]

Answer:

1) MgO + 2HBr -> MgBr2+H2O

2)2N2+5O2->2N2O5

3)C2H4+2O2->CO2+2H2O

4)Fe+2HCl->FeCl2+H2

7 0
3 years ago
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