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8_murik_8 [283]
3 years ago
8

which of these solutions has the lowest freezing point? 0.25 m nacl 0.5 m nacl 1.0 m nacl 1.5 m nacl 2.0 m nacl

Chemistry
2 answers:
zloy xaker [14]3 years ago
6 0

Answer:

2.0 m nacl

Explanation:

Depression of freezing point is given by the equation:

ΔTf = m. Kf. i

where i is the Van't Hoff factor, m is the molality, and Kf is the freezing point depression constant.  

As per the formula, depression of freezing point is directly proportional to the molality. 2.0 m NaCl has the highest molaity and will have the maximum depression in freezing point. As  a result it will also have the lowest freezing point.  

Alinara [238K]3 years ago
4 0
The more particles (ions or molecules) that you can put into solution, the lower the freezing point. 

the answer is E. 2.0 M nacl
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Explanation:

Equation of the reaction:

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The enthalpy change for this reaction will be equal to twice the standard enthalpy change of formation for bromine monochloride, BrCl.

The standard enthalpy change of formation for a compound,

ΔH°f, is the change in enthalpy when one mole of that compound is formed from its constituent elements in their standard state at a pressure of 1 atm.

This means that the standard enthalpy change of formation will correspond to the change in enthalpy associated with this reaction

1/2Br2(g) + 1/2Cl2(g) → BrCl(g)

Here, ΔH°rxn = ΔH°f

This means that the enthalpy change for this reaction will be twice the value of ΔH°f = 2 moles BrCl

Using Hess' law,

ΔH°f = total energy of reactant - total energy of product

= (1/2 * (+112) + 1/2 * (+121)) - 14.7

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What is the theoretical yield of aluminum oxide if 1.40 mol of aluminum metal is exposed to 1.35 mol of oxygen?
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Answer:

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Explanation:

We are given;

  • Moles of Aluminium is 1.40 mol
  • Moles of Oxygen 1.35 mol

We are required to determine the theoretical yield of Aluminium oxide

The equation for the reaction between Aluminium and Oxygen is given by;

4Al(s) + 3O₂(g) → 2Al₂O₃(s)

From the equation 4 moles Al reacts with 3 moles of oxygen to yield 2 moles of Aluminium oxide.

Therefore;

1.4 moles of Al will require 1.05 moles (1.4 × 3/4) of oxygen

1.35 moles of Oxygen will require 1.8 moles (1.35 × 4/3) of Aluminium

Therefore, Aluminium is the rate limiting reagent in the reaction while Oxygen is the excess reactant.

4 moles of aluminium reacts to generate 2 moles aluminium oxide.

Therefore;

Mole ratio Al : Al₂O₃ is 4 : 2

Thus;

Moles of Al₂O₃ = Moles of Al × 0.5

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But; 1 mole of Al₂O₃ = 101.96 g/mol

Thus;

Theoretical mass of Al₂O₃ = 0.7 moles × 101.96 g/mol

                                            = 71.372 g

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3 years ago
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