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)
Oxidation : 3 x (2 I⁻(aq) → I₂(s) + 2 e-)
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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.
<span> In this question (t½) of C-14 is 5730 years, which means that after 5730 years half of the sample would have decayed and half would be left as it is. After 5730 years ( first half life) 70 /2 = 35 mg decays and 35 g remains left. After another 5730 years ( two half lives or 11460 years) 35 /2 = 17.5mg decays and ...</span>Missing: <span>25g</span>
Wouldn't it be 100? 150 - 50 would equal 100. Unless I'm not seeing something.
Anything that has mass and takes up space
For the answer to the question above, one of the most important risks to take into account regarding the mining of uranium is the possibility of nuclear decay taking place and contaminating the worker. So I believe the answer is the last one which is <span>the health hazards that are associated with uranium mining</span>