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Eva8 [605]
3 years ago
15

Many hospitals, and some doctors\' offices, use radioisotopes for diagnosis and treatment, or in palliative care (relief of symp

toms such as pain). Some radioisotopes used in medicine are listed below. Write the isotope symbol for each radioisotope. Replace the question marks with the proper integers. Replace the letter X with the proper element symbol.a) Iodine-131:b) Iridium-192:c) Samarium-153:
Chemistry
1 answer:
jekas [21]3 years ago
7 0

<u>Answer:</u>

<u>For a:</u> The isotopic symbol of the above atom will be _{53}^{131}\textrm{I}

<u>For b:</u> The isotopic symbol of the above atom will be _{77}^{192}\textrm{Ir}

<u>For c:</u> the isotopic symbol of the above atom will be _{62}^{153}\textrm{Sm}

<u>Explanation:</u>

The isotopic representation of an atom is: _Z^A\textrm{X}

where,

Z = Atomic number of the atom

A = Mass number of the atom

X = Symbol of the atom

  • <u>For a:</u>  Iodine-131

Atomic number of iodine = 53

Mass number of iodine = 131

Symbol of iodine = I

The isotopic symbol of the above atom will be _{53}^{131}\textrm{I}

  • <u>For b:</u>  Iridium-192

Atomic number of iridium = 77

Mass number of iridium = 192

Symbol of iridium = Ir

The isotopic symbol of the above atom will be _{77}^{192}\textrm{Ir}

  • <u>For c:</u>  Samarium-153

Atomic number of samarium = 62

Mass number of samarium = 153

Symbol of samarium = Sm

The isotopic symbol of the above atom will be _{62}^{153}\textrm{Sm}

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If 11.7 g of aluminum reacts with 37.2 g of copper (II) sulfate according to the following reaction, how many grams of aluminum
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<u>Step 1:</u> Data given

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<u />

<u>Step 2</u>: The balanced equation

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<u>Step 3</u>: Calculate moles of Aluminium

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<u>Step 4</u>: Calculate moles of CuSO4

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<u>Step 5:</u> Calculate limiting reactant

For 2 moles of Al we need 3 moles of CuSO4

CuSO4 is the limiting reactant. It will be completely consumed (0.233 moles).

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Mass of Al2(SO4)3 = moles Al2(SO4)3 * molar mass Al2(SO4)3

Mass of Al2(SO4)3 = 0.0777 moles * 342.15g/mol

Mass of Al2(SO4)3 = 26.59 grams

There is 26.59 grams of aluminium sulfate produced

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