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(a) Oxygen was formed at the anode and copper was formed at the cathode.
(i) The ionic half-equation for the formation of oxygen is shown.
4OH- = O2 + 2H2O + 4e-
Explain why this reaction is oxidation.
(i) The ionic half-equation for the formation of oxygen is shown.
4OH- = O2 + 2H2O + 4e-
Explain why this reaction is oxidation.
A. Use the Oxidation number method to balance the equation below.
MnO4-(aq) + H+ (aq) + Fe^2+ (aq) = Mn^2+ (aq) + Fe^3+ (aq) + H2O (l)
B. Which element was oxidized and which was reduced?
Oxidized:
Reduced:
MnO4-(aq) + H+ (aq) + Fe^2+ (aq) = Mn^2+ (aq) + Fe^3+ (aq) + H2O (l)
B. Which element was oxidized and which was reduced?
Oxidized:
Reduced:
The oxidation states of vanadium in its compounds are V (+5), V (+4), V (+3) and V (+2). The vanadium (lll) ion can behave as a reductant or an oxidant.
(i) indicate on the following equation which reactant is the oxidant.
2V^3+ + Zn = 2V^2+ + Zn^2+
(i) indicate on the following equation which reactant is the oxidant.
2V^3+ + Zn = 2V^2+ + Zn^2+
The concentration of iron in a solution was found to be 9.9ppm determined by icp-aes.The solution was prepared by solving X gr of FeSO4*7H20(mr=278.01) in 2ml conc,HCLand diluted to 25ml with MQ water in a flask.From primary solution ,5ml was transferred to a 50ml flask and diluted to mark with MQ water and iron conc, was measured from an ordinary matrix matched calibration curve.What was the weight of substance from the beginning if the assay on the can was noted to be 93,5% and what was the molar conc,of sulfate in solution?
hi could you please help me with this ive tried to solve it so far no answer
Why wouldn’t you use an Ultra High Purity chemical to make up a reagent that is used in a colorimetric analysis where the results are in the range of 0 – 40 mg/L with an uncertainty of ± 2 mg/L?
Why wouldn’t you use an Ultra High Purity chemical to make up a reagent that is used in a colorimetric analysis where the results are in the range of 0 – 40 mg/L with an uncertainty of ± 2 mg/L?
1.Oxygen was formed at the anode and copper was formed at the cathode.
(i) The ionic half-equation for the formation of oxygen is shown.
40H- = O2 + 2H2O + 4e-
Explain why this reaction is oxidation.
(ii) Write the ionic half-equation for the formation of copper at the cathode.
(2) The electrolysis was repeated using copper electrodes in place of carbon electrodes.
State and explain what happens to the masses of the anode and the cathode during this electrolysis.
(i) The ionic half-equation for the formation of oxygen is shown.
40H- = O2 + 2H2O + 4e-
Explain why this reaction is oxidation.
(ii) Write the ionic half-equation for the formation of copper at the cathode.
(2) The electrolysis was repeated using copper electrodes in place of carbon electrodes.
State and explain what happens to the masses of the anode and the cathode during this electrolysis.
1. Network addresses are associated with network devices through several different methods identify and describe the common methods.
2. Determine if these devices are on the same subnet or different subnets. Use the address and mask of each address belongs. Show all working outs.
Device A: 192.36.37.50/21
Device B: 192.36.39.55/21
2. Determine if these devices are on the same subnet or different subnets. Use the address and mask of each address belongs. Show all working outs.
Device A: 192.36.37.50/21
Device B: 192.36.39.55/21
1.The oxidation states of vanadium in its compounds are V (+5), V (+4), V (+3) and V (+2). The vanadium (lll) ion can behave as a reductant or an oxidant.
(i) indicate on the following equation which reactant is the oxidant.
2V^3+ + Zn = 2V^2+ + Zn^2+
(ii) Which change in the following equation is oxidation?
Explain your choice.
V^3+ + Fe^3+ = V^4+ + Fe^2+
(i) indicate on the following equation which reactant is the oxidant.
2V^3+ + Zn = 2V^2+ + Zn^2+
(ii) Which change in the following equation is oxidation?
Explain your choice.
V^3+ + Fe^3+ = V^4+ + Fe^2+
1) in the reaction between iodine and thiosulphate ions, the two half reactions are
I2 + 2e- = 2I-
2S2O3^2- = S4O6^2- + 2e-
Use the half equation above to write a single balanced equation, Step by step.
I2 + 2e- = 2I-
2S2O3^2- = S4O6^2- + 2e-
Use the half equation above to write a single balanced equation, Step by step.
(a) Oxygen was formed at the anode and copper was formed at the cathode.
(i) The ionic half-equation for the formation of oxygen is shown.
4OH- = O2 + 2H2O + 4e-
Explain why this reaction is oxidation.
(ii) Write the ionic half-equation for the formation of copper at the cathode.
(b) The electrolysis was repeated using copper electrodes in place of carbon electrodes.
State and explain what happens to the masses of the anode and the cathode during this electrolysis.
(i) The ionic half-equation for the formation of oxygen is shown.
4OH- = O2 + 2H2O + 4e-
Explain why this reaction is oxidation.
(ii) Write the ionic half-equation for the formation of copper at the cathode.
(b) The electrolysis was repeated using copper electrodes in place of carbon electrodes.
State and explain what happens to the masses of the anode and the cathode during this electrolysis.
