Answer:
F2(g)/HF(aq)>Co3 (aq)/Co2 (aq)> H2O2(aq)/H2O(l)> PbO2(s)/Pb(s)>Br2(l)/Br-(aq)
Explanation:
The tendency of any specie to function as oxidizing agent is a highly dependent on the reduction potential of the couple. The more positive the value of the reduction potential of the couple, the better it does as an oxidizing agent.
This implies that we could know a good oxidizing agent by looking at their respective reduction potentials. The couple having the greatest (most positive) reduction potential is selected as the best oxidizing agent. If there are a number of couples at having different reduction potentials, the order of oxidizing ability can be obtained by arranging the species in order of decreasing positive reduction potentials just as we have done in the answer above.
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Answer:
Option D. A liquid with three unsaturated fatty acids.
Explanation:
A careful observation of the molecule given in question shows that the molecule is unsaturated as all three fatty acid that make up the compound contain double.
Double is a characteristic property of unsaturated compound.
Answer:
We need 41.2 L of propane
Explanation:
Step 1: Data given
volume of H2O = 165 L
Step 2: The balanced equation
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Step 3: Calculate moles of H2O
1 mol = 22.4 L
165 L = 7.37 moles
Step 4: Calculate moles of propane
For 1 mol C3H8 we need 5 moles O2 to produce 3 moles CO2 and 4 moles H2O
For 7.37 moles H2O we need 7.37/4 = 1.84 moles propane
Step 5: Calculate volume of propane
1 mol = 22.4 L
1.84 moles = 41.2 L
We need 41.2 L of propane
Answer:
Take 3 mL of the 5 M NaCl solution, 10 mL of the 10% glucose solution, and add water for a final volume of 100 mL.
Explanation:
- In order to calculate the required volume of the 5 M NaCl solution, we calculated the moles contained in a 100 mL solution that has a concentration of 150 mM:
0.1 L * 0.150 M = 0.015 moles of NaCl
With those moles we can calculated the required volume, using the concentration of the stock solution:
0.015 mol / 5 M = 0.003 L = 3 mL.
- To make a solution that has a 1 % concentration of glucose, from a 10 % glucose solution, is the same as to make it ten times less concentrated. Thus, with a final volume of 100 mL, you would need to take 10 mL of the 10% glucose solution, because 100mL * 10/100 = 10.
So in order to prepare the solution, you would need to take 3 mL of the 5 M NaCl solution, 10 mL of the 10% glucose solution, and add water for a final volume of 100 mL.