Showing posts with label CAPE Chemistry Answers. Show all posts
Showing posts with label CAPE Chemistry Answers. Show all posts

Wednesday, December 15, 2021

CAPE Chemistry June 2021 Unit 1 Paper 2 - Past Paper Solutions

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Notes relevant to each question follow the solutions.


Question 1 Notes:

1(e)    It could be that CXC has obtained permission to use past exam questions from         
          different examining bodies. Here is the source of this         




Source Of CAPE Chemistry June 2021 U1 P2 Q1(e)

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The question below was seen before, in a 2005 paper from another examining body.

Its mark scheme follows. The only significant difference is "NaoH" in part (iv) of the CAPE exam vs. "NaOH" in the corresponding part of the original question.













Wednesday, January 10, 2018

CAPE Chemistry Past Paper Answer - 2016 Unit 2 Paper 2 Question 5

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5(a)(i)

As applied to completely miscible binary systems, this states that the partial pressure of a constituent of a binary mixture at any given temperature is equal to the normal vapor pressure of that constituent at the stated temperature multiplied by the mole fraction of the constituent in the mixture.

5(a)(ii)

1. The interactions between components are similar to those in pure components.

2. There is no volume change on mixing the components.

5(b)(i)

An azeotropic mixture is one which boils or distils without change in composition, and in general it has a boiling point higher or lower than that of any of its pure constituents.

5(b)(ii)

An azeotrope is not a compound because its composition varies with pressure.

5(b)(iii)



Mixture X boils at a temperature T1 and at equilibrium gives off a vapor of composition Y richer in the more volatile component A. Continued distillation produces the azeotrope as the distillate while pure B is the residue.

5(c)

At equilibrium, mass of compound in water = (5-x) g and mass of compound in solvent = x g

Partition coefficient = concentration in water/concentration in ether, therefore  ((5-x)/100)/(x/25) = 0.2 i.e. x = 2.8 g.






Wednesday, January 4, 2017

Why Warm Water Freezes Faster Than Cold

Nearly 50 years ago, Erasto B. Mpemba and Denis G. Osborne reported that if samples of water at 90 °C and 25 °C are cooled, the one starting at 90 °C begins freezing firstMany explanations for the “Mpemba effect” have been proposed, including ones based on evaporation, temperature gradients, impurities, and dissolved gases.


In warm water, weak hydrogen bonds break (top, red squiggles), leaving fragments
that easily reorganize into an ice lattice (bottom), a new study says.

A new computational study suggests that the effect arises from the liquid’s hydrogen bond network (J. Chem. Theory Comput. 2016, DOI: 10.1021/acs.jctc.6b00735). Southern Methodist University’s Dieter Cremer and colleagues investigated clusters of 50 and 1,000 water molecules, characterizing the types and strengths of the clusters’ 350 and more than 1 million hydrogen bonds, respectively. In (H2O)1,000 , raising the temperature from 10 °C to 90 °C led to fewer hydrogen bonds, as weaker, predominately electrostatic bonds broke.

That left behind cluster fragments with strong hydrogen bonds with more covalent character and proportionately more “dangling” or terminal hydrogen bonds. That hydrogen bond combination enables the fragments to easily reorganize and form the hexagonal lattice of ice.

Apart from learning what the name of this effect is and why it occurs, you can now answer two unit 1 past paper questions with the knowledge that:

1. Hydrogen bonds are largely electrostatic in nature.


2. Each water molecule forms four hydrogen bonds (from diagram).

Saturday, January 23, 2016

Electronic Configuration – d-block Elements and Ions

The electronic configurations of the first row d-block elements are given in the table below, along with those for the M2+ and M3+ ions. Because the 3d orbitals are all of the same energy, they are each filled with a single electron first.  Only after each of the five 3d orbitals is singly filled (3d5), do the electrons start pairing up, from 3d6 onwards.


For a free atom, the 4s orbital is normally filled before the 3d orbitals. This means that most first row d-block elements have the electronic configuration 3dn 4s2. Chromium (3d5  4s1) and copper (3d10  4s1) are exceptions to this.

The 4s orbital is more diffuse than the 3d orbitals and is affected more by the presence of other atoms or by the charge on the metal. As a result, in an ion or a compound, the 4s orbital is higher in energy than the 3d orbitals. This means that, when electrons are lost to form ions, it is the 4s electrons that are lost first. This is reflected in the electronic configurations of the d-block elements in ions and compounds , as these never contain 4s electrons  unless the d orbitals are full. For example, the electronic configuration of V2+ is 3d3 not 3d1 4s2, and the electronic configuration of Cr(0) in a compound is 3d6, not 3d5 4s1 as it is in atomic chromium.

IMPORTANT: In a compound of a first row d-block element, the 4s orbital is higher in energy than the 3d orbitals. For oxidation states of +2 and higher, the electronic configuration can be found by removing the 4s electrons, plus the appropriate number of 3d electrons, but for lower oxidation states the 4s electrons must be transferred into 3d orbitals before removing electrons.

The video below gives examples of how to obtain electronic configurations for:
Fe3+,
Fe(0), which is different from atomic iron,
Ni2+,
Ni(0),
Co, and,
Co+.