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3 years ago

How many lone pairs of electrons are on the central carbon atom in a Lewis Structure of CHI3?

Chemistry

1 answer:

Alborosie3 years ago

5 0

There are zero lone pairs of electrons on the central carbon atom in a Lewis Structure of CHI3. The correct option is b.

Explanation:

Lone pair electrons are those valence electrons of an atom which do not take part in bonding or left after the bonds are formed.

CH $I_{3}$

carbon is the central atom in the compound having 4 valence electrons.

Each iodine molecule has 7 valence electrons.

H has 1 valence electron.

from the Lewis structure of CH $I_{3}$ , it is seen that all four valence electrons of the carbon is involved in bonding with 3 atoms of iodine and 1 atom of hydrogen.

Thus when all valence electrons are involved in bonding no lone pair of electron is found in CH $I_{3}$ that means zero lone pairs of electrons.

The octet of the iodine, carbon and hydrogen gets completed to form the compound.

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Answer:

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3 years ago

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luda_lava [24]

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2 years ago

How do the resonance structures for ozone, 03, differ?

Anni [7]

Explanation:

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3 years ago

A car with a mass of 1,500 kg is traveling at a speed of 30 m/s. What force must be applied to stop the car in 3 seconds ?

saw5 [17]

Yo sup??

we can solve this problem by applying Newton's 2nd law

F*t=Δp

p=momentum

pi=mu=1500*30

pf=mv=m*0=0

Therefore

F*3=1500*30

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Hope this helps.

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3 years ago

Un móvil se mueve con movimiento acelerado. En los segundo 2 y 3 los espacios recorridos son 90 y 120 m, Calcula la velocidad in

faust18 [17]

Answer:

La velocidad inicial es 55 $\frac{m}{s}$ y su aceleración es -10 $\frac{m}{s^{2} }$

Explanation:

Un movimiento es rectilíneo uniformemente variado, cuando la trayectoria del móvil es una línea recta y su velocidad varia la misma cantidad en cada unidad de tiempo . Dicho de otra manera, este movimiento se caracteriza por una trayectoria que es una línea recta y la velocidad cambia su módulo de manera uniforme: aumenta o disminuye en la misma cantidad por cada unidad de tiempo. Y la aceleración es constante y no nula (diferente de cero).

En este caso la posición del objeto esta dada por la expresión:

$x=x0+v0*t+\frac{1}{2} *a*t^{2}$

donde x es la posición del cuerpo en un instante dado, x0 la posición en el instante inicial, v0 la velocidad inicial y a la aceleración.

En este caso, por un lado podes considerar:

x= 90 m
x0= 0 m
v0= ?
t= 2
a= ?

Reemplazando obtenes:

$90=v0*2+\frac{1}{2} *a*2^{2}$

$90=v0*2+\frac{1}{2} *a*4$

$90=v0*2+2*a$

Y por otro lado tenes:

x= 120 m
x0= 0
v0= ?
t= 3
a= ?

Reemplazando obtenes:

$120=v0*3+\frac{1}{2} *a*3^{2}$

$120=v0*3+\frac{1}{2} *a*9$

$120=v0*3+\frac{9}{2} *a$

Por lo que tenes el siguiente sistema de ecuaciones:

$\left \{ {{2*v0+2*a=90} \atop {3*v0+\frac{9}{2} *a=120}} \right.$

Resolviendo por el método de sustitución, que consiste en aislar en una ecuación una de las dos incógnitas para sustituirla en la otra ecuación, obtenes:

Despejando v0 de la primera ecuación:

$v0= \frac{90-2*a}{2}$