Answer:
74.0 g/mol
Explanation:
Step 1: Write the generic neutralization reaction
HA + NaOH ⇒ NaA + H₂O
Step 2: Calculate the reacting moles of NaOH
At the equivalence point, 33.83 mL of 0.115 M NaOH react.
0.03383 L × 0.115 mol/L = 3.89 × 10⁻³ mol
Step 3: Calculate the moles of HA that completely react with 3.89 × 10⁻³ moles of NaOH
The molar ratio of HA to NaOH is 1:1. The reacting moles of HA is 1/1 × 3.89 × 10⁻³ mol = 3.89 × 10⁻³ mol.
Step 4: Calculate the molar mass of the acid
3.89 × 10⁻³ moles of HA have a mass of 0.288 g.
M = 0.288 g / 3.89 × 10⁻³ mol = 74.0 g/mol
Answer:
See explanation
Explanation:
Anode;
Sn(s) ------> Sn^2+(aq) + 2e
Cathode;
Mn^2+(aq) + 2e ------> Mn(s)
The minimum voltage required to drive the reaction is the cell voltage. The cell voltage is obtained from;
E°cell= E°cathode - E°anode
E°cell= -1.19 - (-0.14)
E°cell= -1.05 V
Answer:
emf generated by cell is 2.32 V
Explanation:
Oxidation: 
Reduction: 
---------------------------------------------------------------------------------
Overall: 
Nernst equation for this cell reaction at 
-
![E_{cell}=E_{cell}^{0}-\frac{0.059}{n}log{[Al^{3+}]^{2}[I^{-}]^{6}}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE_%7Bcell%7D%5E%7B0%7D-%5Cfrac%7B0.059%7D%7Bn%7Dlog%7B%5BAl%5E%7B3%2B%7D%5D%5E%7B2%7D%5BI%5E%7B-%7D%5D%5E%7B6%7D%7D)
where n is number of electrons exchanged during cell reaction, 
is standard cell emf , 
is cell emf , ![[Al^{3+}]](https://tex.z-dn.net/?f=%5BAl%5E%7B3%2B%7D%5D)
is concentration of 
and ![[Cl^{-}]](https://tex.z-dn.net/?f=%5BCl%5E%7B-%7D%5D)
is concentration of 
Plug in all the given values in the above equation -
![E_{cell}=2.20-\frac{0.059}{6}log[(4.5\times 10^{-3})^{2}\times (0.15)^{6}]V](https://tex.z-dn.net/?f=E_%7Bcell%7D%3D2.20-%5Cfrac%7B0.059%7D%7B6%7Dlog%5B%284.5%5Ctimes%2010%5E%7B-3%7D%29%5E%7B2%7D%5Ctimes%20%280.15%29%5E%7B6%7D%5DV)
So, 
Answer: -
0.1 ml of bleach should be added to each liter of test solution.
Explanation:-
Let the volume of bleach to be added is B ml.
Density of stock solution = 1.0 g/ml
Mass of stock solution = Volume of stock x density of stock
= B ml x 1.0 g/ml
= B g
Amount of NaOCl in this stock solution = 5% of B g
= 
x B g
= 0.05 B g
Now each test solution must be added 5 mg/l NaOCl.
Thus each liter of test solution must have 5 mg.
Thus 0.05 B g = 5 mg
= 0.005 g
B = 
= 0.1
Thus 0.1 ml of bleach should be added to each liter of test solution.
When a capacitor has a potential difference between the plates it is said to be Constant.
Both plates have different charge which signifies that one has higher potential than the other.
Therefore, when we join them in parallel, charge will flow from higher to lower. and it continued to flow until equilibrium (the entire process took only a few seconds), indicating that the potential remains constant.
To learn more about constant potential difference in capacitor visit:
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