Answer:
(70 x 2 + 16 x 3) x 29/70 x 2 = 38,9428571 g.
Explanation:
Answer:
Molecular formula
Explanation:
Molecular formula in the first place is required to understand which compound we have. We then should refer to the periodic table and find the molecular weight for each atom. Adding individual molecular weights together would yield the molar mass of a compound.
Then, dividing the total molar mass of a specific atom by the molar mass of a compound and converting into percentage will provide us with the percentage of that specific atom.
E. g., calculate the percent composition of water:
- molecular formula is
; - calculate its molar mass: [tex]M = 2M_H + M_O = 2\cdot 1.00784 g/mol + 16.00 g/mol = 18.016 g/mol;
- find the percentage of hydrogen: [tex]\omega_H = \frac{2\cdot 1.00784 g/mol}{18.016 g/mol}\cdot 100 \% = 11.19 %;
- find the percentage of oxygen: [tex]\omega_O = \frac{16.00 g/mol}{18.016 g/mol}\cdot 100 \% = 88.81 %.
The freezing point depression is a colligative property which means that it is proportional to the number of particles dissolved.
The number of particles dissolved depends on the dissociation constant of the solutes, when theyt are ionic substances.
If you have equal concentrations of two solutions on of which is of a ionic compound and the other not, then the ionic soluton will contain more particles (ions) and so its freezing point will decrease more (will be lower at end).
In this way you can compare the freezing points of solutions of KCl, Ch3OH, Ba(OH)2, and CH3COOH, which have the same concentration.
As I explained the solution that produces more ions will exhibit the greates depression of the freezing point, leading to the lowest freezing point.
In this case, Ba(OH)2 will produce 3 iones, while KCl will produce 2, CH3OH will not dissociate into ions, and CH3COOH will have a low dissociation constant.
Answer: Then, you can predict that Ba(OH)2 solution has the lowest freezing point.
Answer:
The correct answer to you your question is Potassium
Explanation:
Potassium is the most reactive
