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natali 33 [55]
3 years ago
13

The vitamin A molecule has the formula C20H30O, and a molecule of vitamin A2 has the formula C20H28O. Deter- mine how many moles

of vitamin A2 contain the same number of atoms as 1.000 mol vitamin A.
Chemistry
1 answer:
natta225 [31]3 years ago
4 0

Answer:

1.04 moles of vitamin A2 contain the same number of atoms as 1.000 mol vitamin A

Explanation:

Step 1: Data given

Vitamin A = C20H30O

Vitamin A2 = C20H28O

Step 2: Determine the atoms in vitamin A

Vitamin A has 20 C atoms and 30 H atoms and 1 O-atom

In total we have 51 atoms in vitamin A

Step 3: Calculate numbers of atom in 1 mol

1 mol contains 6.022 * 10^23

For 51 atoms = 51 * 6.022*10^23 = 306 *10^23

Step 4: Determine atoms in vitamin A2

Vitamin A2 has 20 C atoms and 28 H atoms and 1 O-atom

In total we have 49 atoms in vitamin A2

Step 5: Calculate molecules A2

Molecules A2 = 306 *10^23  / 49

molecules A2 = 6.24*10^23

Step 6: Calculate how many moles of vitamin A2 contain the same number of atoms as 1.000 mol vitamin A.

n = 6.24*10^23 / 6.022*10^23 = 1.04 mol

1.04 moles of vitamin A2 contain the same number of atoms as 1.000 mol vitamin A

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For doping silicon with boron, silicon specimen was kept in gaseous atmosphere containing B2O3 that maintained the B concentrati
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Solution :

From Fick's law:

$\frac{D_{AB}}{\Delta z } \times (C_{A1}-C_{A2})=N_a$

Mass balance: Exits = Accumulation

-N_A A = \frac{dm}{dt}

-N_A A = \frac{dVp}{dt}

-N_A A = \frac{dV}{dt}p

-N_A A = \frac{dhA}{dt}

-N_A A = \frac{dh}{dt} \times Ap

From the last step, area cancels out and thus leaves :

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So now we can substitute the $N_A$ by the Fick's law

$\frac{D_{AB}}{\Delta z } \times (C_{A1}-C_{A2})=\frac{dh}{dt} p$

Substituting the values we get

$=\frac{-4 \times 10^{-13}}{0.0001} \times (3 \times 10^{26} - C_{A2}) = \frac{dh}{dt} \times 2.46$

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$= -7800 \times 4 \times 10^{-9}  \times (3 \times 10^{26}-C_{A2})=0.000246$$=-(3 \times 10^{26}-C_{A2}) = 7.8846 \times 100 \times \frac{1}{69.62} \times 6.022 \times 10^{23}$

$C_{A2} = 6.85 \times 10^{28} \ \text{ boron atoms} /m^3$

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3 years ago
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