Answer: (1)CaSO4 -> (2)O2 + (1)CaS
Explanation: edge 2020 chem
Answer:
Ka = 6.02x10⁻⁶
Explanation:
The equilibrium that takes place is:
We <u>calculate [H⁺] from the pH</u>:
- [H⁺] =
Keep in mind that [H⁺]=[A⁻].
As for [HA], we know the acid is 0.66% dissociated, in other words:
We <u>calculate [HA]</u>:
Finally we <u>calculate the Ka</u>:
- Ka =
![\frac{[9.12x10^{-4}]*[9.12x10^{-4}]}{[0.138]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B9.12x10%5E%7B-4%7D%5D%2A%5B9.12x10%5E%7B-4%7D%5D%7D%7B%5B0.138%5D%7D)
= 6.02x10⁻⁶
Answer:
First, balance the half-reactions
Second, equalize the electrons
Third,add two reaction equations to get final answer
Explanation:
For example
H₂C₂0₄ + MnO⁻₄ ---------->CO₂+Mn²⁺
(i) Balancing the half reactions
H₂C₂O₄-------->2CO₂+2H⁺+2e⁻
5e⁻ +8H⁺+MnO₄⁻----------->Mn²⁺+4H₂O
(ii)
Equalizing the electrons
5H₂C₂O₄--------->10CO₂+10H⁺+10e⁻ ---here there is a factor of 5
10e⁻+16H⁺+2MnO₄⁻--------->2Mn²⁺+8H₂O -----here there is a factor of 2
(iii)
Add the two where electrons and some Hydrogen ions will cancel out
5H₂C₂O₄+6H⁺+2MnO₄⁻---->10CO₂+2Mn²⁺+8H₂O
Answer:
E° = 1.24 V
Explanation:
Let's consider the following galvanic cell: Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)
According to this notation, Fe is in the anode (where oxidation occurs) and Ag is in the cathode (where reduction occurs). The corresponding half-reactions are:
Anode: Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻
Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an
E° = 0.80 V - (-0.44 V) = 1.24 V
Answer:
The correct answer is "obligatory water reabsorption in the proximal convoluted tubule".
Explanation:
The mechanism for producing concentrated urine cannot include the obligatory reabsorption of water in the proximal convoluted tubule since this process is part of the nephron, the system that filters the blood. Glucose and amino acids are reabsorbed almost entirely, as are approximately 70% of filtered potassium and 80% of bicarbonate.
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