Change one of the variables and try again
Answer:
HI(aq) + H₂O(ℓ) ⟶ H₃O⁺(aq) + I⁻(aq)
Explanation:
The HI donates a proton to the water, converting it to a hydronium ion
HI(aq) + H₂O(ℓ) ⟶ H₃O⁺(aq) + I⁻(aq)
Thus, the HI is behaving like a Brønsted acid.
Answer:
At anode - 
At cathode - 
Explanation:
Electrolysis of NaBr:
Water will exist as:
The salt, NaBr will dissociate as:
At the anode, oxidation takes place, as shown below.
At the cathode, reduction takes place, as shown below.
Answer: The image from the question has the correct answers.
Explanation:
As summarized in the attached table.
Answer:
The answer to your question is P = 0.18 atm
Explanation:
Data
mass of O₂ = 0.29 g
Volume = 2.3 l
Pressure = ?
Temperature = 9°C
constant of ideal gases = 0.082 atm l/mol°K
Process
1.- Convert the mass of O₂ to moles
16 g of O₂ -------------------- 1 mol
0.29 g of O₂ ---------------- x
x = (0.29 x 1)/16
x = 0.29/16
x = 0.018 moles
2.- Convert the temperature to °K
Temperature = 9 + 273 = 282°K
3.- Use the ideal gas law ro find the answer
PV = nRT
-Solve for P
P = nRT/V
-Substitution
P = (0.018 x 0.082 x 282) / 2.3
-Simplification
P = 0.416/2.3
-Result
P = 0.18 atm
