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The type of substances that would make good conductors would be substances that has greater density. This would mean that the particles of the substance are much closer to each other and would allow the free flow of the electrons in the substance. With this, current would be able to flow freely.
Volume of base(NaOH)= Vb =5 mL
Concentration of base(NaOH)= Mb =2 M
Volume of acid(HCl)= Va =10 mL
Concentration of acid(HCl)= Ma = ?
Now,
Ma*Va = Mb*Vb
Ma*(10) = (2)*(5)
Ma= 1M
Answer:
All three are present
Explanation:
Addition of 6 M HCl would form precipitates of all the three cations, since the chlorides of these cations are insoluble:
.
- Firstly, the solid produced is partially soluble in hot water. Remember that out of all the three solids, lead(II) choride is the most soluble. It would easily completely dissolve in hot water. This is how we separate it from the remaining precipitate. Therefore, we know that we have lead(II) cations present, as the two remaining chlorides are insoluble even at high temperatures.
- Secondly, addition of liquid ammonia would form a precipitate with silver:
; Silver hydroxide at higher temperatures decomposes into black silver oxide:
. - Thirdly, we also know we have
in the mixture, since addition of potassium chromate produces a yellow precipitate:
. The latter precipitate is yellow.
Answer:
144 g
Explanation:
Use the mole ratio of 4 mol CO2 for every 9 mol O2 to convert from mol O2 to mol CO2. Then use the molar mass of CO2 to convert from mol of CO2 to grams of CO2.
7.34 mol O2 • (4 mol CO2 / 9 mol O2) • (44.01 g CO2 / 1 mol CO2) = 144 g CO2