Correct answer is option E. <span>It is a redox reaction in which Zn is oxidized at the anode, and V is reduced at the cathode.
Reason:
In above reaction, the oxidation state of VO3- is +5, while that of VO2 is +4. Thus there is reduction of V from +5 to +4
In case of Zn, oxidation state of Zn is increased from 0 to +2, Thus process is referred as oxidation. </span>
Answer:
Solubility of O₂(g) in 4L water = 3.42 x 10⁻² grams O₂(g)
Explanation:
Graham's Law => Solubility(S) ∝ Applied Pressure(P) => S =k·P
Given P = 0.209Atm & k = 1.28 x 10⁻³mol/L·Atm
=> S = k·P = (1.28 x 10⁻³ mole/L·Atm)0.209Atm = 2.68 x 10⁻³ mol O₂/L water.
∴Solubility of O₂(g) in 4L water at 0.209Atm = (2.68 x 10⁻³mole O₂(g)/L)(4L)(32 g O₂(g)/mol O₂(g)) = <u>3.45 x 10⁻² grams O₂(g) in 4L water. </u>
Answer:
d. 8 moles of H2O on the product side
Explanation:
Hello,
In this case, we need to balance the given redox reaction in acidic media as shown below:
![MnO_4^{1-} (aq) + Cl^{1-} (aq) \rightarrow Mn^{2+} (aq) + Cl_2 (g)\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq) + Cl^{1-} (aq) \rightarrow Mn^{2+} (aq) + Cl_2 (g)\\\\\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O\\\\2Cl^{1-}\rightarrow Cl_2^0+2e^-\\\\2*[(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O]\\\\5*[2Cl^{1-}\rightarrow Cl_2^0+2e^-]\\\\\\\\2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10e^- \rightarrow 2Mn^{2+}+8H_2O\\\\10Cl^{1-}\rightarrow 5Cl_2^0+10e^-\\](https://tex.z-dn.net/?f=MnO_4%5E%7B1-%7D%20%28aq%29%20%2B%20Cl%5E%7B1-%7D%20%28aq%29%20%5Crightarrow%20%20Mn%5E%7B2%2B%7D%20%28aq%29%20%2B%20Cl_2%20%28g%29%5C%5C%5C%5C%28Mn%5E%7B7%2B%7DO%5E%7B2-%7D_4%29%5E%7B1-%7D%20%28aq%29%20%2B%20Cl%5E%7B1-%7D%20%28aq%29%20%5Crightarrow%20%20Mn%5E%7B2%2B%7D%20%28aq%29%20%2B%20Cl_2%20%28g%29%5C%5C%5C%5C%5C%5C%5C%5C%28Mn%5E%7B7%2B%7DO%5E%7B2-%7D_4%29%5E%7B1-%7D%20%28aq%29%2B8H%5E%2B%2B5e%5E-%20%5Crightarrow%20Mn%5E%7B2%2B%7D%2B4H_2O%5C%5C%5C%5C2Cl%5E%7B1-%7D%5Crightarrow%20Cl_2%5E0%2B2e%5E-%5C%5C%5C%5C2%2A%5B%28Mn%5E%7B7%2B%7DO%5E%7B2-%7D_4%29%5E%7B1-%7D%20%28aq%29%2B8H%5E%2B%2B5e%5E-%20%5Crightarrow%20Mn%5E%7B2%2B%7D%2B4H_2O%5D%5C%5C%5C%5C5%2A%5B2Cl%5E%7B1-%7D%5Crightarrow%20Cl_2%5E0%2B2e%5E-%5D%5C%5C%5C%5C%5C%5C%5C%5C2%28Mn%5E%7B7%2B%7DO%5E%7B2-%7D_4%29%5E%7B1-%7D%20%28aq%29%2B16H%5E%2B%2B10e%5E-%20%5Crightarrow%202Mn%5E%7B2%2B%7D%2B8H_2O%5C%5C%5C%5C10Cl%5E%7B1-%7D%5Crightarrow%205Cl_2%5E0%2B10e%5E-%5C%5C)
Then, we add the half reactions:
Thereby, we can see d. 8 moles of H2O on the product side.
Best regards.
Question 4 is <span>A. longitudinal waves </span>
