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The element magnesium, Mg, proton number 12, is a metal which is used in many alloys which are strong and light - CIE - A-Level Chemistry - Question 1 - 2010 - Paper 1

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The element magnesium, Mg, proton number 12, is a metal which is used in many alloys which are strong and light. Magnesium has several naturally occurring isotopes. ... show full transcript

Worked Solution & Example Answer:The element magnesium, Mg, proton number 12, is a metal which is used in many alloys which are strong and light - CIE - A-Level Chemistry - Question 1 - 2010 - Paper 1

Step 1

What is meant by the term isotope?

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Answer

An isotope refers to atoms of the same element that have the same number of protons (atomic number) but different numbers of neutrons. This results in different mass numbers for the isotopes.

Step 2

Complete the table below for two of the isotopes of magnesium.

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Answer

IsotopeNumber of ProtonsNumber of NeutronsNumber of Electrons
24Mg121212
26Mg121412

Step 3

Calculate the relative atomic mass, Ar, of magnesium in the sample.

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Answer

To calculate the relative atomic mass, we use the formula:

Ar=(24×78.60)+(25×10.11)+(26×11.29)100A_r = \frac{(24 \times 78.60) + (25 \times 10.11) + (26 \times 11.29)}{100}

Calculating each term separately:

  • For 24Mg: 24×78.60=1896.4024 \times 78.60 = 1896.40
  • For 25Mg: 25×10.11=252.7525 \times 10.11 = 252.75
  • For 26Mg: 26×11.29=293.5426 \times 11.29 = 293.54

Adding these together gives:

Ar=1896.40+252.75+293.54100=1868.69A_r = \frac{1896.40 + 252.75 + 293.54}{100} = 1868.69

Thus, the relative atomic mass of magnesium in the sample, rounded to four significant figures, is 24.33.24.33.

Step 4

Construct a balanced equation for the reaction between magnesium and chlorine.

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Answer

The balanced equation for the reaction between magnesium and chlorine is:

Mg+Cl2MgCl2Mg + Cl_2 \rightarrow MgCl_2

Step 5

Calculate the amount, in moles, of antimony atoms that reacted.

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Answer

To calculate the amount of antimony in moles:

n(Sb)=massmolar mass=2.45122=0.0200n(Sb) = \frac{mass}{molar\ mass} = \frac{2.45}{122} = 0.0200

Step 6

Calculate the amount, in moles, of chlorine atoms that reacted.

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Answer

From the mass of the chloride A formed:

  1. Calculate the mass of chlorine: mass of Cl=mass of Amass of Sbmass\ of\ Cl = mass\ of\ A - mass\ of\ Sb mass of Cl=4.572.45=2.12gmass\ of\ Cl = 4.57 - 2.45 = 2.12 g

  2. Using molar mass of Cl (35.5 g/mol): n(Cl)=2.1235.5=0.0597n(Cl) = \frac{2.12}{35.5} = 0.0597

Step 7

Use your answers to (i) and (ii) to determine the empirical formula of A.

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Answer

Using the mole ratios calculated:

  • Moles of Sb = 0.0200
  • Moles of Cl = 0.0597

The ratio is approximately: Sb:Cl=0.0200:0.05971:3Sb : Cl = 0.0200 : 0.0597 \approx 1 : 3 Thus, the empirical formula of A would be SbCl3SbCl_3.

Step 8

Construct a balanced equation for the reaction between antimony and chlorine.

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Answer

The balanced equation for the reaction between antimony and chlorine is:

2Sb+3Cl22SbCl32Sb + 3Cl_2 \rightarrow 2SbCl_3

Step 9

What type of bonding is present in magnesium chloride?

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Answer

The type of bonding present in magnesium chloride is ionic bonding.

Step 10

Suggest what type of bonding is present in A.

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Answer

The type of bonding present in compound A (SbCl3) is covalent bonding.

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