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kiruha [24]
2 years ago
14

How many molecules of CO2 will be produced by the decomposition

Chemistry
1 answer:
34kurt2 years ago
4 0

Answer:

2.99×10²⁵ molecules of CO₂ are produced

Explanation:

Decomposition reaction is:

Ca(HCO₃)₂ => CaO(s) + 2CO₂(g) + H₂O(g)

Ratio is 1:2. Let's make a rule of three:

1 mol of bicarbonate can produce 2 moles of CO₂

Therefore, 24.9 moles of bicarbonate may produce, 49.8 moles (24.9 .2 )/1

Let's determine the number of molecules

1 mol has 6.02×10²³ molecules

49.8 moles must have (49.8 . 6.02×10²³) / 1 = 2.99×10²⁵ molecules

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3 years ago
Choose the aqueous solution below with the lowest freezing point. These are all solutions of nonvolatile solutes and you should
atroni [7]

Answer:

(NH_4)_3PO_4 0.075 m solution has the lowest freezing point.

Explanation:

Depression in freezing point is given by:

\Delta T_f=T-T_f

\Delta T_f=K_f\times m

\Delta T_f=iK_f\times \frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in Kg}}

where,

\Delta T_f =Depression in freezing point

K_f = Freezing point constant of solvent

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m = molality

According question, molality of all the solutions are same and are in prepared with same solvent. So, values of molality and K_f will remain the same and will not effect the freezing point of the solution.

The lowering in freezing point will now depend upon van't Hoff factors of the solutions. Higher the value of van'Hoff factor more will be the lowering in freezing point of the solution.

The van't Hoff factor of KNO_2 solution = i_1=2

The van't Hoff factor of LiCN solution = i_2=2

The van't Hoff factor of (NH_4)_3PO_4 solution = i_3=4

The van't Hoff factor of NaI solution = i_4=2

The van't Hoff factor of NaBrO_3 solution = i_5=2

The solution of ammonium phosphate has the highest values of van't Hoff factor which will result in maximum lowering of the freezing point of the solution.

Hence,(NH_4)_3PO_4 0.075 m solution has the lowest freezing point.

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