Answer:
1610.7 g is the weigh for 4.64×10²⁴ atoms of Bi
Explanation:
Let's do the required conversions:
1 mol of atoms has 6.02×10²³ atoms
Bi → 1 mol of bismuth weighs 208.98 grams
Let's do the rules of three:
6.02×10²³ atoms are the amount of 1 mol of Bi
4.64×10²⁴ atoms are contained in (4.64×10²⁴ . 1) /6.02×10²³ = 7.71 moles
1 mol of Bi weighs 208.98 g
7.71 moles of Bi must weigh (7.71 . 208.98 ) /1 = 1610.7 g
Answer:
Option c → Tert-butanol
Explanation:
To solve this, you have to apply the concept of colligative property. In this case, freezing point depression.
The formula is:
ΔT = Kf . m . i
When we add particles of a certain solute, temperature of freezing of a solution will be lower thant the pure solvent.
i = Van't Hoff factor (ions particles that are dissolved in the solution)
At this case, the solute is nonvolatile, so i values 1.
ΔT = Difference between fussion T° of pure solvent - fussion T° of solution.
T° fussion paradichlorobenzene = 56 °C
T° fussion water = 0°
T° fussion tert-butanol = 25°
Water has the lowest fussion temperature and the paradichlorobenzene has the highest Kf. But the the terbutanol, has the highest Kf so this solvent will have the largest change in freezing point, when all the molalities are the same.
Answer: B
Explanation: molarity = concentration c= n/V = 0.5 mol/ 0.05 l = 10 mol/l
yeag
Explanation:
2SrO + 4NO2 + O. The thermal decomposition of strontium nitrate to produce strontium oxide, nitrogen dioxide and oxygen. This reaction takes place at a temperature of over 570°C
