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Bogdan [553]
3 years ago
12

The purging of impurities from a compound by crystallization:

Chemistry
2 answers:
Nuetrik [128]3 years ago
5 0
C.

This is also the only change by which crystals may be produced.
Ne4ueva [31]3 years ago
3 0
<h3><u>Answer;</u></h3>

occurs after the liquid melt cools back to a solid

The purging of impurities from a compound by crystallization <u>occurs after the liquid melt cools back to a solid.</u>

<h3><u>Explanation;</u></h3>
  • <em><u>Purging of impurities from a compound by crystallization is done through a method known as melt crystallization. Melt crystallization is a type of cooling crystallization process which is operated close to the melting point of the pure compound.</u></em>
  • The sample for melt crystallization id normally an impure compound. When the impure sample is cooled below the equilibrium temperature the result is formation of a solid phase which is more pure that the initial compound, while the impurities remain in the impure sample.
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Iridium has only two naturally occurring isotopes. The mass of iridium-191 is 190.9605 amu and the mass of iridium-193 is 192.96
cluponka [151]

Answer:

Abundance of Iridium-193 is 62.75%

Explanation:

From the question given above, the following data were obtained:

Isotope A (Iridium-191):

Mass of A = 190.9605 amu

Abundance of A = A%

Isotope B (Iridium-193):

Mass of B = 192.9629 amu

Abundance B = (100 – A) %

Relative atomic mass of Iridium = 192.217 amu

Next, we shall determine the abundance of isotope A (Iridium-191). This can be obtained as follow:

Relative atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100]

192.217 = [(190.9605 × A%)/100] + [(192.9629 × (100 – A)%)/100]

192.217 = 1.909605A% + 1.929629(100 – A)%

192.217 = 1.909605A% + 192.9629 – 1.929629A%

Collect like terms

192.217 – 192.9629 = 1.909605A% – 1.929629A%

–0.7459 = –0.020024A%

Divide both side by –0.020024

A% = –0.7459 / –0.020024

A% = 37.25 %

Finally, we shall determine the abundance of Isotope B (Iridium-193).

This can be obtained as follow:

Abundance of A (Iridium-191) = 37.25 %

Abundance of B (Iridium-193) =?

Abundance B = 100 – A%

Abundance B = 100 – 37.25 %

Abundance of B (Iridium-193) = 62.75%

4 0
3 years ago
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Answer:

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Answer:

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Explanation:

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Write the full ionic equation and net ionic equation for sodium dihydrogen phosphate + calcium carbonate, sodium oxilate + calcl
My name is Ann [436]

Answer:

<em>Sodium dihydrogen phosphate + calcium carbonate</em>

<u>Full ionic equation</u>

2 Na⁺(aq) + 2 H₂PO₄⁻(aq) + CaCO₃(s) ⇄ 2 Na⁺(aq) + CO₃²⁻(aq) + Ca(H₂PO₄)₂(s)

<u>Net ionic equation</u>

2 H₂PO₄⁻(aq) + CaCO₃(s) ⇄ CO₃²⁻(aq) + Ca(H₂PO₄)₂(s)

<em>Sodium oxalate + calcium carbonate</em>

<u>Full ionic equation</u>

2 Na⁺(aq) + C₂O₄²⁻(aq) + CaCO₃(s) ⇄ 2 Na⁺(aq) + CO₃²⁻(aq) + CaC₂O₄(s)

<u>Net ionic equation</u>

C₂O₄²⁻(aq) + CaCO₃(s) ⇄ CO₃²⁻(aq) + CaC₂O₄(s)

<em>Sodium hydrogen phosphate + calcium carbonate</em>

<u>Full ionic equation</u>

2 Na⁺(aq) + HPO₄²⁻(aq) + CaCO₃(s) ⇄ CaHPO₄(s) + 2 Na⁺(aq) + CO₃²⁻(aq)

<u>Net ionic equation</u>

HPO₄²⁻(aq) + CaCO₃(s) ⇄ CaHPO₄(s) + CO₃²⁻(aq)

Explanation:

Let's consider two kind of equations:

  • Full ionic equation: includes all ions and species that do not dissociate in water.
  • Net ionic equation: includes only ions that participate in the reaction (<em>not spectator ions</em>) and species that do not dissociate in water.
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