What is [H+] given that the measured cell potential is -0.464 V and the anode reduction ... What half-reaction occurs at the cathode during the electrolysis of molten ... PbO2(s) + 4H+(aq) + SO42-(aq) + 2e- → PbSO4(s) + 2H2O(l); E° = 1.69 V .... For the cell Cu(s)|Cu2+||Ag+|Ag(s), the standard cell potential is 0.46 V. A cell ... hopw this helps
<span>The oxygen atom accepts the proton. The oxidation number of O is -2, meaning that there are two unshared electrons in the valence shell; In the ClO- ion, one of these is shared with the Cl- ion, leaving an unshared electron on the oxygen atom, which is what the hydrogen atom shares its electron with, becoming the proton accepted by the O atom.</span>
Answer:
-2
Explanation:
7 x 1 - 2 x 1 + 1 x 1 + 3C = 0 (no charge)
6 + 3C = 0
C = -2
Answer:
ATP supply will reduced or decrease drastically
Explanation:
ATP will reduced because carbohydrates are energy given food. Chemical energy are in carbohydrates, therefore when the undergo digestion i.e breakdown of larger pieces into smaller pieces, it is broken down into glucose i.e energy is released. Therefore, if there is now carbohydrates, ATP supply will reduced and there won't be energy needed for different cellular metabolic activities. The breakdown of carbohydrate supply energy.
Answer:
M of HI = 5.4 M.
Explanation:
- We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.
<em>(XMV) acid = (XMV) base.</em>
where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.
M is the molarity of the acid or base.
V is the volume of the acid or base.
<em>(XMV) HI = (XMV) Ca(OH)₂.</em>
For HI; X = 1, M = ??? M, V = 25.0 mL.
For Ca(OH)₂, X = 2, M = 1.5 M, V = 45.0 mL.
<em>∴ M of HI = (XMV) Ca(OH)₂ / (XV) HI</em> = (2)(1.5 M)(45.0 mL) / (1)(25.0 mL) = <em>5.4 M.</em>
