Changes

| style="height:20px; width:300px; text-align:center;" |Find Calculate the '''specific charge of a [[proton]].| style="height:20px; width:300px; text-align:center;" |Find Calculate the '''specific charge''' of a [[Magnesium|Magnesium-24]] 2+ [[ion]].

<math>Q_{proton} Q = 1\times1.60\times10^{-19}Coulombs</math>[[Coulomb|C]]

<math>Mass_{proton} m = 1\times1.67\times10^{-27}kilograms</math>[[Kilogram|kg]]

<math>Q_{Mg+2} Q = 2\times1.60\times10^{-19}Coulombs</math>[[Coulomb|C]]

<math>Mass_{Mg-24} m = (24\times1.67\times10^{-27}kilograms + 10\times9.11\times10^{-31})</math>[[Kilogram|kg]]

<math>S.C. = \frac{2\times1.60\times10^{-19}}{24\times1.67\times10^{-27}kilograms + 10\times9.11\times10^{-31}}</math>

| style="height:20px; width:200px300px; text-align:center;" |Calculate the [[Electrical Charge|charge]] of an a [[Cobalt]] [[ion]] with a '''specific charge''' of 4.... 87x10⁶Ckg^-1 and a [[mass]] of 9....86x10^-26kg.| style="height:20px; width:200px300px; text-align:center;" |Calculate the [[Electrical Charge|charge]] of a [[Atomic Nucleus|nucleus]] with a '''specific charge''' of 4.... 51x10⁷Ckg^-1 and a [[mass]] of 8...... 52x10^-26kg and identify the [[element]].

| style="height:20px; width:200px300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s.''' <math>S.C._{Co} = 4.87\times10^{6}</math>Ckg^-1 <math>m_{Co} = 9.86\times10^{-26}</math>[[Kilogram|kg]] | style="height:20px; width:200px300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s.''' <math>S.C. = 4.51\times10^{7}</math>Ckg^-1 <math>m = 8.52\times10^{-26}</math>[[Kilogram|kg]] 

| style="height:20px; width:200px300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].''' <math>S.C. = \frac{Q}{m}</math> <math>4.87\times10^{6} = \frac{Q}{9.86\times10^{-26}}</math> | style="height:20px; width:200px300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].''' <math>S.C. = \frac{Q}{m}</math> <math>4.51\times10^{7} = \frac{Q}{8.52\times10^{-26}}</math> 

| style="height:20px; width:200px300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].''' <math>Q = 4.87\times10^{6} \times 9.86\times10^{-26}</math> <math>Q = 4.80\times10^{-19}</math>[[Coulomb|C]] | style="height:20px; width:200px300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].''' <math>Q = 4.51\times10^{7} \times 8.52\times10^{-26}</math> <math>Q = 3.84\times10^{-18}</math>[[Coulomb|C]] To identify the [[Atomic Nucleus|nucleus]] divide the [[Electrical Charge|charge]] by the [[Elementary Charge|elementary charge]] to find the number of [[proton]]s. <math>Z = \frac{3.84\times10^{-18}}{1.60\times10^{-19}}</math> <math>Z = 24.0</math> There are 24 [[proton]]s in the [[Atomic Nucleus|nucleus]] therefore the [[element]] is [[Chromium]].

| style="height:20px; width:200px300px; text-align:center;" |Calculate the [[mass]] of a an [[particleion]] with an a [[Electrical Charge|charge]] of -3.... 20x10^-19[[Coulomb|C]] and a '''specific charge''' of -5....80x10⁶Ckg^-1.| style="height:20px; width:200px300px; text-align:center;" |Calculate the [[Relative Atomic Mass|relative atomic mass]] of a [[particleAtomic Nucleus|nucleus]] with an [[Electrical Charge|charge]] of 9.... 61x10^-19[[Coulomb|C]] and a '''specific charge''' of 4....11x10⁷Ckg^-1.

| style="height:20px; width:200px300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s.''' <math>S.C. = -5.80\times10^{6}</math>Ckg^-1 <math>Q = -3.20\times10^{-19}</math>[[Coulomb|C]]| style="height:20px; width:200px300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s.''' <math>S.C. = 4.11\times10^{7}</math>Ckg^-1 <math>Q = 9.61\times10^{-19}</math>[[Coulomb|C]]

 | style="height:20px; width:200px300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].''' <math>S.C. = \frac{Q}{m}</math> <math>-5.80\times10^{6} = \frac{-3.20\times10^{-19}}{m}</math>| style="height:20px; width:200px300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].''' <math>S.C. = \frac{Q}{m}</math> <math>4.11\times10^{7} = \frac{9.61\times10^{-19}}{m}</math>

 | style="height:20px; width:200px300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].''' <math>m=\frac{-3.20\times10^{-19}}{-5.80\times10^{6}}</math> <math>m=5.52\times10^{-26}kg</math>| style="height:20px; width:200px300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].'''|}<math>m====Calculating Atomic Mass====\frac{9.61\times10^{-19}}{4.11\times10^{7}}</math>{| class="wikitable"| style<math>m="height:20px; width:200px; text2.34\times10^{-align:center;" |Calculate 26}kg</math> To find the [[Atomic Mass|atomic mass]] of the divide this [[Atomic Nucleus|nucleus]] of an [[isotope]] of [[Leadmass]] with a '''specific charge''' of ........| style="height:20px; width:200px; text-align:center;" |Calculate by the [[Atomic Mass|atomic mass]] of the a single [[Atomic Nucleus|nucleus]] of an [[isotope]] of [[Iron]] with the '''specific charge''' of ........ for an [[Iron|Iron 3+]] [[ionnucleon]].|-| style<math>A="height:20px; width:200px; text-align:left;" |'''1\frac{2. State the known quantities in [[SI Unit]]s.'''| style="height:20px; width:200px; text34\times10^{-align:left;" |'''26}}{1. State the known quantities in [[SI Unit]]s.'''|67\times10^{-27}}</math>| style="height:20px; width:200px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].'''| style<math>A="height:20px; width:200px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]]14.'''0</math>|-| style="height:20px; width:200px; text-align:left;" |'''3. The [[Rearrange (Maths)Relative Atomic Mass|Rearrange]] the equation and [[Solve (Maths)|solverelative atomic mass]]is 14.'''| style="height:20px; width:200px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].'''

| style="height:20px; width:200px300px; text-align:center;" |Calculate the number of [[neutron]]s in an [[isotope]] of [[Titanium]] with a [[Atomic Nucleus|nucleus]] of '''specific charge''' of ......4.58x10⁷Ckg^-1.| style="height:20px; width:200px300px; text-align:center;" |Calculate the number of [[neutron]]s in an [[isotope]] of [[Neon]] with the '''specific charge''' of -2....... 46x10⁶Ckg^-1 for a [[NeonSelenium|Neon Selenium 2-]] [[ion]].

| style="height:20px; width:200px300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s.''' <math>S.C. = 4.58\times10^{7}</math>Ckg^-1 There are 22 [[proton]]s in the [[Atomic Nucleus|nucleus]] of [[Titanium]]. <math>Q = 22\times1.60\times10^{-19}</math>C <math>m = (z + n)\times1.67\times10^{-27}</math>kg Where z is the number of [[proton]]s and n is the number of [[neutron]]s and (z+n) is the [[Relative Atomic Mass|relative atomic mass]]. Therefore <math>m = (22 + n)\times1.67\times10^{-27}</math>kg | style="height:20px; width:200px300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s.''' <math>S.C. = -2.46\times10^{6}</math>Ckg^-1 A 2- [[ion]] has two more [[electron]]s than [[proton]]s giving it a [[Negative Charge|negative charge]]. <math>Q = -2\times1.60\times10^{-19}</math>C <math>m = (z + n)\times1.67\times10^{-27}+(z+2)\times9.11\times10^{-31}</math>kg Where z is the number of [[proton]]s, n is the number of [[neutron]]s, (z+n) is the [[Relative Atomic Mass|relative atomic mass]] and (z+2) is the number of [[electron]]s since there are two more [[electron]]s than [[proton]]s. Therefore <math>m = (34 + n)\times1.67\times10^{-27}+36\times9.11\times10^{-31}</math>kg 

| style="height:20px; width:200px300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].''' <math>S.C. = \frac{Q}{m}</math> <math>4.58\times10^{7} = \frac{22\times1.60\times10^{-19}}{(22 + n)\times1.67\times10^{-27}}</math> | style="height:20px; width:200px300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].''' <math>S.C. = \frac{Q}{m}</math> <math>-2.46\times10^{6} = \frac{-2\times1.60\times10^{-19}}{(34 + n)\times1.67\times10^{-27}+36\times9.11\times10^{-31}}</math> 

| style="height:20px; width:200px300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] and [[Evaluate (Maths)|evaluate]] the equation and , then [[Solve (Maths)|solve]].''' <math>(22 + n) = \frac{22\times1.60\times10^{-19}}{1.67\times10^{-27}\times4.58\times10^{7}}</math> <math>n = \frac{22\times1.60\times10^{-19}}{1.67\times10^{-27}\times4.58\times10^{7}} - 22</math> <math>n = 46.0 - 22</math> <math>n = 24.0</math> There are 24 [[neutron]]s in this isotope of [[Titanium]].| style="height:20px; width:200px300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] and [[Evaluate (Maths)|evaluate]] the equation and , then [[Solve (Maths)|solve]].''' <math>(34 + n)\times1.67\times10^{-27}+36\times9.11\times10^{-31} = \frac{-2\times1.60\times10^{-19}}{-2.46\times10^{6}}</math> <math>(34 + n)\times1.67\times10^{-27}+3.2796\times10^{-29} = 1.3008\times10^{-25}</math> <math>(34 + n) = \frac{1.3008\times10^{-25}-3.2796\times10^{-29}}{1.67\times10^{-27}}</math> <math>(34 + n) = 77.9</math> <math>n = 43.9</math>The [[Atomic Nucleus|nucleus]] contains 44 [[neutron]]s.

| style="height:20px; width:200px300px; text-align:center;" |Calculate the number of [[proton]]s in a [[Atomic Nucleus|atomic nucleus]] of an [[isotope]] with a an [[Relative Atomic Mass|atomic mass]] of ..... 231[[Unified Atomic Mass Units|u]] and a [[Specific Charge|specific charge]] of 3......82x10⁷Ckg^-1.| style="height:20px; width:200px300px; text-align:center;" |Calculate the number of [[proton]]s in a [[Atomic Nucleus|atomic nucleus]] of an [[isotope]] with a 30 [[massneutron]] of ..... s and the a [[Specific Charge|specific charge]] of 4....... for its 5+ [[ion]]35x10⁷Ckg^-1.

| style="height:20px; width:200px300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s.''' <math>S.C. = 3.82\times10^{7}</math>Ckg^-1 <math>Q = z\times1.60\times10^{-19}</math>C <math>m = 231\times1.67\times10^{-27}</math>kg| style="height:20px; width:200px300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s.''' <math>S.C. = 4.35\times10^{7}</math>Ckg^-1 <math>Q = z\times1.60\times10^{-19}</math>C The [[Relative Atomic Mass|atomic mass]] can be approximated as the number of [[nucleon]]s multiplied by the [[Unified Atomic Mass Units|unified atomic mass unit]]. <math>m = (z+30)\times1.67\times10^{-27}</math>kg

| style="height:20px; width:200px300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].''' <math>S.C. = \frac{Q}{m}</math> <math>3.82\times10^{7} = \frac{z\times1.60\times10^{-19}}{231\times1.67\times10^{-27}}</math> | style="height:20px; width:200px300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].''' <math>S.C. = \frac{Q}{m}</math> <math>4.35\times10^{7} = \frac{z\times1.60\times10^{-19}}{(z+30)\times1.67\times10^{-27}}</math>

| style="height:20px; width:200px300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].''' <math>3.82\times10^{7}\times231\times1.67\times10^{-27} = z\times1.60\times10^{-19}</math> <math>1.4736414\times10^{-17} = z\times1.60\times10^{-19}</math> <math>z = \frac{1.4736414\times10^{-17}}{1.60\times10^{-19}}</math> <math>z = 92.1</math> The [[element]] is [[Uranium]].| style="height:20px; width:200px300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].''' <math>4.35\times10^{7} = \frac{z\times1.60\times10^{-19}}{(z+30)\times1.67\times10^{-27}}</math> <math>4.35\times10^{7}\times(z+30)\times1.67\times10^{-27} = z\times1.60\times10^{-19}</math> <math>(z+30)\times7.2645\times10^{-20} = z\times1.60\times10^{-19}</math> <math>z\times7.2645\times10^{-20} + 30\times7.2645\times10^{-20}= z\times1.60\times10^{-19}</math> <math>z\times1.60\times10^{-19} - z\times7.2645\times10^{-20} = 30\times7.2645\times10^{-20}</math> <math>z(1.60\times10^{-19} - 7.2645\times10^{-20}) = 30\times7.2645\times10^{-20}</math> <math>z = \frac{30\times7.2645\times10^{-20}}{1.60\times10^{-19} - 7.2645\times10^{-20}}</math> <math>z = 24.9</math> There are 25 [[proton]]s so the [[element]] is [[Manganese]].