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Strike441 [17]
3 years ago
6

Which condition increases the number of collisions between reactant molecules in a given volume?

Chemistry
2 answers:
Nezavi [6.7K]3 years ago
4 0

Answer:

Increasing Surface Area

Explanation:

A greater surface area (meaning more, smaller particles) allows for more opportunity for particles to collide.  On the other hand, decreasing temperature and removing a catalyst would only decrease the number of collisions, and the clumping option doesn't make much sense.  Hope this helps!

Nata [24]3 years ago
3 0

Answer:

Increasing Surface Area

Explanation:

Chemical reactions occur at the atomic level where the atoms of each reacting elements interact with one another to form products. However, one way this interaction occurs is via COLLISION OF ATOMIC PARTICLES. In accordance with the collision theory, a collision must occur between reactant atomic particles in order for a reaction to take place.

However, certain factors affect the rate at which this collision occurs. One of them is the SURFACE AREA OF THE REACTANTS. Increasing the surface area of reactants simultaneously increases the frequency at which particles collide. Hence, INCREASING SURFACE AREA increases the number of collisions between reactant molecules in a given volume.

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What is the silver ion concentration in a solution prepared by mixing 369 mL 0.373 M silver nitrate with 411 mL 0.401 M sodium c
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Answer:

[A g + ]  =  3.12 *10^-6 M

Explanation:

Step 1: Data given

Volume silver nitrate = 369 mL

Molarity silver nitrate = 0.373 M

Volume sodium chromate = 411 mL

Molarity sodium chromate = 0.401 M

The Ksp of silver chromate is 1.2 * 10^− 12

Step 2: The balanced equation

2 A g + ( a q )  +  C rO4^2-  ( a q )  →Ag2CrO4 (  s)

Step 3:

[Ag+]i = [AgNO3] * V1/(V1+V2) * 1 mol Ag+ / 1 mol AgNO3

[Ag+]i = 0.373 M * 0.369/ (0.369+0.411)

[Ag+]i = 0.176 M

[C rO4^2]i = [Ag2CrO4] * V2 /(V1+V2) * 1mol C rO4^2 / 1 mol Ag2CrO4

[C rO4^2i = 0.401 M * 0.411 / (0.369+0.411)

[C rO4^2]i = 0.211 M

Ksp = [Ag+]²[CrO4^2-] = 1.2 * 10^− 12

[CrO4^2-]f = [CrO4^2-]i - 0.5 * [Ag+]i

[CrO4^2-]f = 0.211 -0.088 = 0.123 M

1.2 * 10^− 12  = ( 2 x ) ²*( 0.123M + x )

[ C O 3^ −2] f  >>  x

1.2 * 10^− 12  = ( 2 x ) ²* 0.123M

x = 1.56 * 10^-6

[A g + ] f  = 2x = 3.12 *10^-6 M

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