1) in periodic acid (HIO₄), iodine has oxidation number +7, hydrogen has oxidation number +1, oxygen has -2, compound has neutral charge:
+1 + x + 4 · (-2) = 0.
x = +7.
2) in molecule of iodine (I₂), iodine has oxidation number 0, because iodine is nonpolar molecule.
3) in sodium iodide (NaI), iodine has oxidation number -1, sodium has oxidation number +1:
+1 + x = 0.
x = -1.
4) in iodic acid (HIO₃), iodine has oxidation number +5, hydrogen has oxidation number +1, oxygen has -2, compound has neutral charge:
+1 + x + 3 · (-2) = 0.
x = +5.
Explanation:
Start with a balanced equation.
2H2 + O2 → 2H2O
Assuming that H2 is in excess, multiply the given moles H2O by the mole ratio between O2 and H2O in the balanced equation so that moles H2O cancel.
5 mol H2O × (1 mol O2/2 mol H2O) = 2.5 mol O2
Answer: 2.5 mol O2 are needed to make 5 mol H2O, assuming H2 is in excess.
Answer:
The pH value of the mixture will be 7.00
Explanation:
Mono and disodium hydrogen phosphate mixture act as a buffer to maintain pH value around 7. Henderson–Hasselbalch equation is used to determine the pH value of a buffer mixture, which is mathematically expressed as,
![pH=pK_{a} + log(\frac{[Base]}{[Acid]})](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%20%2B%20log%28%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%29)
According to the given conditions, the equation will become as follow
![pH=pK_{a} + log(\frac{[Na_{2}HPO_{4} ]}{[NaH_{2}PO_{4}]})](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%20%2B%20log%28%5Cfrac%7B%5BNa_%7B2%7DHPO_%7B4%7D%20%5D%7D%7B%5BNaH_%7B2%7DPO_%7B4%7D%5D%7D%29)
The base and acid are assigned by observing the pKa values of both the compounds; smaller value means more acidic. NaH₂PO₄ has a pKa value of 6.86, while Na₂HPO₄ has a pKa value of 12.32 (not given, but it's a constant). Another more easy way is to the count the acidic hydrogen in the molecular formula; the compound with more acidic hydrogens will be assigned acidic and vice versa.
Placing all the given data we obtain,
Given:
Concentration of titrant = 0.1000 M
Volume of titrant = 45 mL
The molarity of analyte depends on the amount of the analyte present in the titrated solution. If the amount of analyte is 20 mL, then its concentration is:
45ml * 0.10 M = C analyte * 20 ml
C analyte = 0.225 M
The answer is true. It is the last star.
