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arsen [322]
3 years ago
5

Many hospitals use radioisotopes for diagnosis and treatment or in palliative care. Three radioisotopes used in medicine are giv

en. Write the isotope symbol for each radioisotope. Replace the question marks with the proper integers. Replace the letter X with the proper element symbol.
Chemistry
1 answer:
ehidna [41]3 years ago
6 0

The question is incomplete, here is the complete question:

Many hospitals use radioisotopes for diagnosis and treatment, or in palliative care (relief of symptoms 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:

<u>Answer:</u>

<u>For a:</u> The isotopic representation of the given isotope is: _{53}^{131}\textrm{I}

<u>For b:</u> The isotopic representation of the given isotope is: _{77}^{192}\textrm{Ir}

<u>For c:</u> The isotopic representation of the given isotope is: _{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

For the given options:

<u>Option a:</u>  Iodine-131

The atomic number of iodine is 53 and the atomic mass of the given isotope is 131. The chemical symbol for iodine element is 'I'

Hence, the isotopic representation of the given isotope is: _{53}^{131}\textrm{I}

<u>Option b:</u>  Iridium-192

The atomic number of iridium is 77 and the atomic mass of the given isotope is 192. The chemical symbol for iridium element is 'Ir'

Hence, the isotopic representation of the given isotope is: _{77}^{192}\textrm{Ir}

<u>Option c:</u>  Samarium-153

The atomic number of samarium is 62 and the atomic mass of the given isotope is 153. The chemical symbol for iridium element is 'Sm'

Hence, the isotopic representation of the given isotope is: _{62}^{153}\textrm{Sm}

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3 years ago
If the fugacity of a pure component at the conditions of an ideal solution is 40 bar and its mole fraction is 0.4, what is its f
yaroslaw [1]

Answer : The fugacity in the solution is, 16 bar.

Explanation : Given,

Fugacity of a pure component = 40 bar

Mole fraction of component  = 0.4

Lewis-Randall rule : It states that in an ideal solution, the fugacity of a component is directly proportional to the mole fraction of the component in the solution.

Now we have to calculate the fugacity in the solution.

Formula used :

f_i=X_i\times f_i^o

where,

f_i = fugacity in the solution

f_i^o = fugacity of a pure component

X_1 = mole fraction of component

Now put all the give values in the above formula, we get:

f_i=0.4\times 40\text{ bar}

f_i=16\text{ bar}

Therefore, the fugacity in the solution is, 16 bar.

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3 years ago
Does density depend on the amount of substance that you have? Explain your answer.
Natasha_Volkova [10]

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Density depends on the amount of the substance you have, as the mass will increase, but also what the volume is because if you have a high mass object with an extremely high volume, it won't be very dense. But if you have a high mass object with a low volume, it will be very dense.

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How many hydrogen atoms are in 35.0 grams of hydrogen gas? How many hydrogen atoms are in 35.0 grams of hydrogen gas? 4.25 × 102
Ede4ka [16]

Answer: 2.12\times 10^{25} atoms of hydrogen are there in

35.0 grams of hydrogen gas.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{35.0g}{2g/mol}=17.5moles

1 mole of hydrogen (H_2) = 2\times 6.023\times 10^{23}=12.05\times 10^{23} atoms

17.5 mole of hydrogen (H_2) = \frac{12.05\times 10^{23}}{1}\times 17.5=2.12\times 10^{25} atoms

There are 2.12\times 10^{25} atoms of hydrogen are there in

35.0 grams of hydrogen gas.

8 0
3 years ago
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