Answer:
pCl⁻ = 0.54
Explanation:
First we <u>calculate how many Cl⁻ moles are coming from each substance</u>, using the <em>given volumes and concentrations</em>:
- 0.12 M NaCl * 0.10 L = 0.012 mol NaCl = 0.012 mol Cl⁻
- 0.18 M MgCl₂ * 0.23 L = 0.0414 mol MgCl₂ = (0.0414 * 2) 0.0828 mol Cl⁻
The final volume of the mixture is = 0.10 L + 0.23 L = 0.33 L
Now we <u>calculate [Cl⁻]</u>, using the<em> total number of Cl⁻ moles and the final volume:</em>
- [Cl⁻] = (0.012 mol + 0.0828 mol) / 0.33 L = 0.29 M
Finally we <u>calculate the pCl⁻ of the resulting solution</u>:
Answer:
electrons
Explanation:
A redox reaction is one in which oxidation and reduction takes place. in the sense of
oxidation being the loss of electrons and reduction the gain of electrons.
The reduced forms of the electron carries NAD+ / NADH and FADH / FADH2 have high potential energy.
- NAD& FAD used to donate electrons as a reducing agent, they receive electrons from other molecules and then became reduced.
-FAD is called flow in adenine dinucleotide, it is a redox cofactor and it is in many important enzymatic reactions in metabolism.
Compounds that has two elements on the formula can be bonded by ionic or covalent bond. If the compound is a metal and a non-metal then it will be ionic bonding. However, when it involves two non-metals, then it will have covalent bonding.
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.
