Yes, that is true. in order for it to be a redox reaction, both oxidation and reduction must be occurring.
Answer:
The concentration of the copper (II) sulfate solution is 2.06 * 10^2 μmol/L or 2.06 * 10^2 μM
Explanation:
The concentration of a solution is the amount of solute dissolved in a given volume of solution. In this case, the concentration of the copper(II) sulfate solution in micromoles per liter (symbol ) is the number of micromoles of copper(II) sulfate dissolved in each liter of solution. To calculate the micromoles of copper(II) sulfate dissolved in each liter of solution you must divide the total micromoles of solute by the number of liters of solution.
Here's that idea written as a formula: c= n/V
where c stands for concentration, n stands for the total micromoles of copper (II) sulfate and V stands for the total volume of the solution.
You're not given the volume of the solution in liters, but rather in milliliters. You can convert milliliters to liters with a unit ratio: V= 150. mL * 10^-3 L/ 1 mL = 0.150 L
Next, plug in μmol and liters into the formula to divide the total micromoles of solute by the number of liters of solution: c= 31 μmol/0.150 L = 206.66 μmol/L
Convert this number into scientific notation: 2.06 * 10^2 μmol/L or 2.06 * 10^2 μM
Answer:
A. Chipping ice to flakes
Explanation:
You can't reverse the ice becoming flakes. They will stay like that until they melt
<span>Answer:
</span><span>
</span><span>
</span><span>Li⁺ (aq) + OH⁻ (aq) + H⁺ (aq) + Cl⁻(aq) → Li⁺ (aq) + Cl⁻ (aq) + H₂O(l)</span><span />
<span>Explanation:
</span>
<span>1) Combine the cation Li⁺ (aq) with the anion Cl- (aq) to form LiCl(s).
</span>
<span>LiCl is a solid soluble substance, a typical ionic compound. So, it will reamain as separate ions in the product side: Li⁺ + CL⁻</span>
<span>2) Combine the anion OH⁻ with the cation H⁺ to form H₂O(l).
</span>
<span>Since, the ionization of H₂O is low, it will remain as liquid in the product side: H₂O(l)</span>
<span>3) Finally, you can wirte the total ionic equation:
</span>
Li⁺ (aq) + OH⁻ (aq) + H⁺ (aq) + Cl⁻(aq) → Li⁺ (aq) + Cl⁻ (aq) + H₂O(l)
