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Sedbober [7]
3 years ago
10

Rare earth elements plz

Chemistry
2 answers:
Slav-nsk [51]3 years ago
3 0
Rare earth metals are a group of 17 elements - lanthanum, cerium, praseodymium, neodymium, promethium, samarium, europium, gadolinium, terbium, dysprosium, holmium, erbium, thulium, ytterbium, lutetium, scandium, yttrium - that appear in low concentrations in the ground
Westkost [7]3 years ago
3 0

Answer:

Rare earth elements are lanthanum, cerium, praseodymium, neodymium, promethium, samarium, europium, gadolinium, terbium, dysprosium, holmium, erbium, thulium, ytterbium, lutetium, scandium, yttrium - that appear in low concentrations in the ground.

Explanation:

The rare earth elements (REE) are a set of seventeen metallic elements. These include the fifteen lanthanides on the periodic table plus scandium and yttrium. Rare earth elements are an essential part of many high-tech devices.

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What is the coefficient for iodide ions (I-) when the equation is balanced?<br>​
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Answer:

When the following redox equation is balanced with smallest whole number coefficients, the coefficient for the iodide ion will be __6__.

Explanation:

From the redox equation, we can see that NO₃⁻ is reduced to NO (from oxidation state +5 to +2), whereas I⁻ is oxidized to I₂ (from oxidation state -1 to 0). The half reactions are balanced with H⁺ (acidic solution), as follows:

Reduction :        2 x (NO₃⁻(aq) + 3 e-  + 4 H⁺ → NO(g) + 2 H₂O)

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                      ----------------------------------------------------------------------

Total equation: 6 I⁻(aq) + 2 NO₃⁻(aq)+ 8 H⁺ → 3 I₂(s) + 2 NO(g) + 4 H₂O

That is the redox equation with the smallest whole number coefficients.

Accordin to this, the coefficient for the iodide ion (I⁻) is: 6.

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