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AveGali [126]
3 years ago
5

Given the propane molecule C3H8(g)! a) What is the balanced equation for the combustion of propane in O2?(20 pts.) b) What is th

e standard enthalpy change ΔH0 for the reaction at 25°C? (20 pts.) c) Finally, is this reaction exothermic or endothermic, under what condition is the answer to part b) the heat of the reaction? (10 pts.)
Chemistry
1 answer:
lubasha [3.4K]3 years ago
4 0

Answer:

The answer to your question is below

Explanation:

Propane = C₃H₈

Process

1.- Write the chemical reaction

                     C₃H₈  +  O₂   ⇒   CO₂  +  H₂O

balanced chemical reaction

                      C₃H₈  +  5O₂   ⇒  3CO₂  +  4H₂O

               Reactants      Elements     Products

                      3                    C                  3

                      8                    H                  8

                     10                    O                 10

b) Standard enthalpy

Propane                  -104.7 kJ/mol

Oxygen                         0   kJ/mol

Carbon dioxide      -393.5 kJ/mol

Water                      -241.8 kJ/mol

ΔH° = ∑ΔH° products - ∑H° reactants

ΔH° = 3(-393.5) + 4(-241.8) - [(-104.7) + 0]

-Simplification

ΔH° = -1180.5 kJ/mol - 967.2 kJ/mol + 104.7 kJ/mol

ΔH° = - 2042.5 kJ/mol

This reaction is exothermic because ΔH° is negative.

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The Xenopus laevis oocyte is roughly 1mm in diameter. After fertilization, it undergoes mitosis, a process requiring the coordin
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Answer:

a)  the diffusion length  = 1000 micrometers;

so if the diffusion co-efficient is also known or given; we can therefore deduce the time taken( i.e  how long it would take for a message to reach from one side of the cell to the other).

b) We need to be given the data value concerning the diffusion coefficient of the protein.

c) 1.234 × 10⁷ seconds

Explanation:

a)

In a controlled transportation that occurs in diffusion; the time taken can be expressed as:

time (t) = \frac{diffusion length^2}{diffusion co-efficient }

Now; if a message is passed through a Ca^{2+} ion, From studies; we know that these Ca^{2+} ions are infinitesimally smaller in size in the Ca^{2+} channels.

Also the information being carried through the Ca^{2+} ions are bound to be transported diagonally across the cell.

However; the diffusion length  = 1000 micrometers;

so if the diffusion co-efficient is also known or given; we can therefore deduce the time taken( i.e  how long it would take for a message to reach from one side of the cell to the other).

b) Since; we don't know the information concerning the diffusion coefficient of the protein, then it is quite not possible to determine how long it would take for a message to be transmitted through the diffusion of a small protein.

c) Here; we are given the value of the diffusion coefficient to be = 0.2 micrometer²/seconds

In a transmembrane protein; the diffusion length is half the circumference of the cell which is = \pi*r

= \pi * \frac{1000}{2}

= 1570.8 micrometers

Now;the time taken can now be calculated as:

time (t) = \frac{diffusion length^2}{diffusion co-efficient }

time (t) = \frac{(1570.8)^2}{0.2 }

time (t) =1.234*10^7 seconds

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3 years ago
What occurs during one half-life?
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4 years ago
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Si tenemos una solución de ácido láctico 0.025M con un pH = 2.75. ¿Cuál es la constante de equilibrio Ka?
rodikova [14]

Answer:

La constante de equilibrio Ka del ácido láctico es 1.38x10⁻⁴.

Explanation:

El ácido láctico es un ácido débil cuya reacción de disociación es la siguiente:

CH₃CHOHCOOH + H₂O ⇄ CH₃CHOHCOO⁻ + H₃O⁺   (1)

0.025M - x                                        x                     x                    

La constante de acidez del ácido es:        

Ka = \frac{[CH_{3}CHOHCOOH^{-}][H_{3}O^{+}]}{[CH_{3}CHOHCOOH]}

Sabemos que la concentración del ácido inicial es:

[CH₃CHOHCOOH] = 0.025 M    

Y que a partir del pH podemos hallar [H₃O⁺]:

pH = -log[H_{3}O^{+}]

[H_{3}O^{+}] = 10^{-pH} = 10^{-2.75} = 1.78 \cdo 10^{-3} M

Debido a que el ácido se disocia en agua para producir los iones CH₃CHOHCOO⁻ y H₃O⁺ de igual manera (según la reacción (1)), tenemos:

[CH₃CHOHCOO⁻] = [H₃O⁺] = 1.78x10⁻³ M

Y por esa misma disociación, la concentración del ácido en el equilibrio es:

[CH_{3}CHOHCOOH^{-}] = 0.025 M - 1.78 \cdo 10^{-3} M = 0.023 M

Entonces, la constante de equilibrio Ka del ácido láctico es:

Ka = \frac{[CH_{3}CHOHCOOH^{-}][H_{3}O^{+}]}{[CH_{3}CHOHCOOH]} = \frac{(1.78 \cdo 10^{-3})^{2}}{0.023} = 1.38 \cdot 10^{-4}          

 

Espero que te sea de utilidad!  

3 0
3 years ago
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