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2 years ago

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If 0.5 ml of a mumps virus vaccine contains 5000 units of antigen, how many units would be present in each milliliter if the 0.5

ml of vaccine was diluted to 2 ml with water for injection?

1 answer:

kobusy [5.1K]2 years ago

3 0

5000 units of antigens are present in 0.5 mL of the mumps virus vaccine.

Volume of the vaccine after dilution with water = 2 mL

0.5 mL has 5000 units antigen. So when 0.5 mL was diluted to 2 mL, the final vaccine after dilution would contain 5000 units of antigen.

Units of antigen present in 1 mL vaccine sample

= $1 mL *\frac{5000 units}{2 mL} = 2500 units$

Therefore, 2500 units of the antigen will be present in each milliliter, if 0.5 mL of the vaccine is diluted to 2 mL with water for injection.

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(I)how many atoms are present in 7g of lithium?

ICE Princess25 [194]

Answer :

(i) The number of atoms present in 7 g of lithium are, $6.07\times 10^{23}$

(ii) The number of atoms present in 7 g of lithium are, $1.204\times 10^{24}$

(iii) The number of moles of $F_2$ is, 1 mole

The number of moles of $CO_2$ is, 0.5 mole

The number of moles of $OH^-$ is, 1 mole

Explanation :

<u>Part (i) :</u>

First we have to calculate the moles of lithium.

$\text{Moles of }Li=\frac{\text{Mass of }Li}{\text{Molar mass of }Li}$

Molar mass of Li = 6.94 g/mole

$\text{Moles of }Li=\frac{7g}{6.94g/mol}=1.008mole$

Now we have to calculate the number of atoms present.

As, 1 mole of lithium contains $6.022\times 10^{23}$ number of atoms

So, 1.008 mole of lithium contains $1.008\times 6.022\times 10^{23}=6.07\times 10^{23}$ number of atoms

Thus, the number of atoms present in 7 g of lithium are, $6.07\times 10^{23}$

<u>Part (ii) :</u>

First we have to calculate the moles of carbon.

$\text{Moles of }C=\frac{\text{Mass of }C}{\text{Molar mass of }C}$

Molar mass of C = 12 g/mole

$\text{Moles of }C=\frac{24g}{12g/mol}=2mole$

Now we have to calculate the number of atoms present.

As, 1 mole of carbon contains $6.022\times 10^{23}$ number of atoms

So, 2 mole of carbon contains $2\times 6.022\times 10^{23}=1.204\times 10^{24}$ number of atoms

Thus, the number of atoms present in 7 g of lithium are, $1.204\times 10^{24}$

<u>Part (iii) :</u>

<u>To calculate the moles of </u> $F_2$ <u> :</u>

$\text{Moles of }F_2=\frac{\text{Mass of }F_2}{\text{Molar mass of }F_2}$

Molar mass of $F_2$ = 38 g/mole

$\text{Moles of }F_2=\frac{19g}{19g/mol}=1mole$

Thus, the number of moles of $F_2$ is, 1 mole

<u>To calculate the moles of </u> $CO_2$ <u> :</u>

$\text{Moles of }CO_2=\frac{\text{Mass of }CO_2}{\text{Molar mass of }CO_2}$

Molar mass of $CO_2$ = 44 g/mole

$\text{Moles of }CO_2=\frac{22g}{44g/mol}=0.5mole$

Thus, the number of moles of $CO_2$ is, 0.5 mole

<u>To calculate the moles of </u> $OH^-$ <u> ions :</u>

$\text{Moles of }OH^-=\frac{\text{Mass of }OH^-}{\text{Molar mass of }OH^-}$

Molar mass of $OH^-$ = 17 g/mole

$\text{Moles of }OH^-=\frac{17g}{17g/mol}=1mole$

Thus, the number of moles of $OH^-$ is, 1 mole

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pychu [463]

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we have to know the spin of valence electrons of carbon-14

There are four unpaired electron which are called as valence electron also.The spin of the four unpaired electron is either upfilled or down filled.

The ground state electronic configuration of C-atom is 1s²2s²2p² and one electron from 2s orbital gets excited to 2p orbital. The elctronic configuration in excited state is 1s²2s¹ $2p_{x}^{1} 2p_{y}^{1}2p_{z}^{1}$ .

The electron jumps because half-filled orbitals are more stable. Exchange energy is less than pairing energy.

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pochemuha

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Explanation:

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