Chemistry, 29.03.2021 18:50 klandry0

What will happen to a soccer ball that is rolling across a field if an unbalanced force of friction is acting on the soccer ball? A. The soccer ball will gradually move slower and slower until it stops.
B. The soccer ball will continue to move at a constant speed.
C. The soccer ball will move at a constant speed for a while, then slow down and stop.
D. The soccer ball will slow down for a while and then move at a constant speed without stopping.

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Answer from: Tink2334

C. The soccer ball will move at a constant speed for a while then slow down and stop

Answer from: Quest

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: 1 - α α α = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

$k_{c}$ = $\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)v }$

or, $k_{c}$ = $\frac{\alpha ^{2} }{(1-\alpha) }$ (at v = 1l)

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: (1 - α)/0.5 α/0.5 α/0.5 = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

$\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)0.5 }$

$\frac{\alpha ^{2} }{(1-\alpha) }$ = $\frac{\alpha ^{2} }{(1-\alpha) 0.5}$

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

Answer from: Quest

For cyclohexane compounds drawn in a chair conformation: -axial bonds are vertical -equatorial bonds are parallel to two bonds within the ring. thus, as i assume the chair conformation is drawn: -the x group is in an axial position, as are the five other vertical bonds. -three axial bonds are on the top face of the ring -the other three axial bonds are on the bottom face the position of the second substituent on each ring carbon is equatorial. gauche groups must be on an adjacent carbon such as carbon 6 and carbon 2, and also gauche groups are rotated by 60 degrees relative to each other. drawing the newman projections for c1–c2 and c1–c6 aids in visualizing the groups that are gauche to the y group on c1. click this link and watch some academic videos that may you more understand the topic

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