Answer:
0.00370 g
Explanation:
From the given information:
To determine the amount of acid remaining using the formula:
where;
v_1 = volume of organic solvent = 20-mL
n = numbers of extractions = 4
v_2 = actual volume of water = 100-mL
k_d = distribution coefficient = 10
∴
Thus, the final amount of acid left in the water = 0.012345 * 0.30
= 0.00370 g
Answer:
A) oxidizing agent is SO2
B) NaClO is the oxidizing agent
Explanation:
A) This is a redox reaction in which oxidation and reduction occur simultaneously.
Thus, in 2H2S(g) + SO2(g) -> 2H2O(l) + 3S(s);
H2S is reduced as follows;
H2S → S + 2H+ + 2e−
We can see that SO2 has been reduced while H2S gets oxidized since it has changed state from - 2 to 0 . Thus sulphur dioxide is the oxidizing agent.
B) SO2(g) + H2O(l) + NaClO(aq) -> NaCl(aq) + H2SO4(aq)
In this, SO2 undergoes oxidation and NaClO is the oxidizing agent
Explanation:
Once solid ammonium nitrate interacts with water, the molecules of polar water intermingle with these ions and attract individual ions from the structure of the lattice, that actually will break down. E.g;-NH4NO3(s) — NH4+(aq)+ NO3-(aq) To split the ionic bonds that bind the lattice intact takes energy that is drained from the surroundings to cool the solution.
Some heat energy is produced once the ammonium and nitrate ions react with the water molecules (exothermic reaction), however this heat is far below that is needed by the H2O molecules to split the powerful ionic bonds in the solid ammonium nitrate.
Hence, we can say that the dissolution of ammonium nitrate in water is highly endothermic reaction.
Answer: emission spectrum
Explanation:
The answer would be millimeters
so answer choice d
