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2 years ago

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A vial containing radioactive selenium-75 has an activity of 3.0 mCi/mL. If 2.6 mCi are required for a leukemia test, how many m

icroliters must be administered?
A vial containing radioactive selenium-75 has an activity of 3.0 mCi/mL. If 2.6 mCi are required for a leukemia test, how many microliters must be administered?

1 answer:

oksian1 [2.3K]2 years ago

4 0

Answer : The $866.66\mu L$ must be administered.

Solution :

As we are given that a vial containing radioactive selenium-75 has an activity of $3.0mCi/mL$ .

As, 3.0 mCi radioactive selenium-75 present in 1 ml

So, 2.6 mCi radioactive selenium-75 present in $\frac{2.6mCi}{3.0mCi}\times 1ml=0.86666ml\times 1000=866.66\mu L$

Conversion :

$(1ml=1000\mu L)$

Therefore, the $866.66\mu L$ must be administered.

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<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

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<u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 9.06 g of carbon dioxide, $\frac{12}{44}\times 9.06=2.47g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 5.58 g of water, $\frac{2}{18}\times 5.58=0.62g$ of hydrogen will be contained.

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To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{2.47g}{12g/mole}=0.206moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.62g}{1g/mole}=0.62moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{3.29g}{16g/mole}=0.206moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.206 moles.

For Carbon = $\frac{0.206}{0.206}=1$

For Hydrogen = $\frac{0.62}{0.206}=3$

For Oxygen = $\frac{0.206}{0.206}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 3 : 1

Hence, the empirical formula for the given compound is $C_1H_{3}O_1=CH_3O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 124 amu = 124 g/mol

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Putting values in above equation, we get:

$n=\frac{124g/mol}{31g/mol}=4$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 4)}H_{(3\times 4)}O_{(1\times 4)}=C_4H_{12}O_4$

Thus, the empirical and molecular formula for the given organic compound is $CH_3O$ and $C_4H_{12}O_4$

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2. Li₂S(aq) + Pb(NO₃)₂(aq) → PbS + 2 Li(NO₃)

3. Pb(ClO₃)₂(aq) + 2 NaNO₃(aq) → Pb(NO₃)₂ + 2 Na(ClO₃)

4. AgNO₃(aq) + KCl(aq) → AgCl + KNO₃

5. 2 K₂S(aq) + Sn(NO₃)₄(aq) → SnS₂ + 4 KNO₃

K₂S is a soluble sulfide that, in solid state, is white. As K has a low electronegativity, the relative polar bond K-S allows its dissolution in water.

PbS is a black and insoluble compound. As Pb electronegativity is higher than K electronegativity, The bond is less-polar and its dissolution will not be allowed.

For the third reaction, all nitrates are soluble and Na(ClO₃) is also a very soluble compound.

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As Sn electronegativity is high, the solubility of this compound is very low. Also, this compound is black!

Thus, combinations could yield a black precipitate are:

Li₂S(aq) + Pb(NO₃)₂(aq)

2 K₂S(aq) + Sn(NO₃)₄(aq)

I hope it helps!

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