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vladimir2022 [97]
3 years ago
6

Which statement best describes the temperature dependence of an addition reaction? Addition reactions are thermodynamically impo

ssible. Addition reactions are thermodynamically disfavored at all temperatures. Addition reactions are thermodynamically favored at low temperatures. Addition reactions are thermodynamically favored at high temperatures. Addition reactions are thermodynamically favored at all temperatures.
Chemistry
2 answers:
Nookie1986 [14]3 years ago
8 0

Answer:

Addition reactions are thermodynamically favored at low temperatures.

Explanation:

kakasveta [241]3 years ago
5 0

Answer:

Addition reactions are thermodynamically favored at low temperatures.

Explanation:

Compared to substitutions or eliminations, addition reactions do not require to break as many bonds as them, as such, they do not require such a high input of energy (ie. temperature) in order to take place.

This is why if there's a high temperature, the reactions that require more energy -like substitutions or eliminations- will be more thermodinamically favored than the reactions that require less energy -like additions-, and viceversa.

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Explanation:

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3 years ago
How are the formulas and names of ionic compounds written
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3 0
3 years ago
For doping silicon with boron, silicon specimen was kept in gaseous atmosphere containing B2O3 that maintained the B concentrati
Kay [80]

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 :

-N_A  = \frac{dh}{dt} \times p

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

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

$=-4 \times 10^{-9} \times( 3 \times 10^{26} - C_{A2}) = 0.0001 \times 2.46$

$= -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}$

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

3 0
3 years ago
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