Oxygen almost always has an oxidation number of -2
The only exception is in H2O2 which makes it -1, and in OF2 which makes it +2
Answer:
5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺
Explanation:
We will balance the redox reaction through the ion-electron method.
Step 1: Identify both half-reactions
Reduction: Br₂ ⇒ Br⁻
Oxidation: S₂O₃²⁻ ⇒ SO₄²⁻
Step 2: Perform the mass balance, adding H⁺ and H₂O where appropriate
Br₂ ⇒ 2 Br⁻
5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺
Step 3: Perform the charge balance, adding electrons where appropriate
2 e⁻ + Br₂ ⇒ 2 Br⁻
5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻
Step 4: Make the number of electrons gained and lost equal
5 × (2 e⁻ + Br₂ ⇒ 2 Br⁻)
1 × (5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻)
Step 5: Add both half-reactions
5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺
Answer:
a. is an excel document. i uploaded as an attachment
b. concentration = 0.327
Explanation:
in the graph i uploaded for answer a, the x variable is conductivity, while the y variable is % salinity.
the calculated regression line was calculated using excel =
y = 0.057655X - 0.00891
for answer part b,
we take x = 5.82
when we put this into the y formular
y = 0.057655(5.82) - 0.00891
y = 0.3355521 - 0.00891
y = 0.3266421
this is ≈ 0.327
in conclusion, using the standard curve, the concentration of unknown salt is 0.327. the % salinity = 0.327
Given:
Sulfur dioxide and dihydrogen sulfide system
Vapor Pressure of water, H2O = 23 torr
P_SO2 = P_H2S
Temperature of the system = 298 K
Assume: Total pressure of the system is 1 atm = 760 torr
If partial pressure of SO2 is equal to the concentration of H2S, then,
760 = 0.5*PSO2 + 0.5PH2S
760 = PSO2
The reaction is
2H₂(g) + O₂(g) ---> 2H₂O
Thus as per balanced equation two moles of hydrogen will react with one moles of oxygen.
There is a directly relation between moles and volume. [One mole of each gas occupies 22.4 L of volume at STP]
Thus we can say that two unit volume of hydrogen will react with one unit volume of oxygen
Now as we have started with equal units of volume of both oxygen and hydrogen, half of oxygen will be consumed against complete volume of hydrogen
so the gas which will remain in excess is oxygen
