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myrzilka [38]
3 years ago
13

Why does a lower concentration of dissolved particles decrease the reaction rate? A. It does not. A high concentration decreases

reaction rate. B. When there are less dissolved particles, less collisions take place. C. Concentration is measured inversely. A low concentration indicates a higher portion of dissolved reactant particles. D. The dissolved reactant cannot collide with enough energy to react because it is moving slowly when it is dissolved in a solution.
Chemistry
1 answer:
Ratling [72]3 years ago
3 0

Answer: Option (B) is the correct answer.

Explanation:

Whenever there is less concentration of solute particles in a solvent then it means less number of solute particles are available. As a result, there will occur less number of collisions between the solvent and solute particles.

It means that there will be a decrease in rate of reaction.

But if there is more concentration of solute particles then it shows more number of solute particles are available for reaction. As a result, more number of collisions will take place between the particles of solute and solvent.

Hence, then there will occur an increase in rate of reaction.

Thus, we can conclude that a lower concentration of dissolved particles decrease the reaction rate because when there are less dissolved particles, less collisions take place.

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Please help!
Dahasolnce [82]

Answer:

reproduction

Explanation:

reproduction, process by which organisms replicate themselves

6 0
3 years ago
Ba(OH)2 Ba+2 + 2 OH- (dissolved in solution). Which will NOT happen to the equilibrium of this solution as H+ ions are added? H+
prohojiy [21]

Answer:

The reaction will move to the left.

Explanation:

  • For the reaction:

<em>Ba(OH)₂ = Ba²⁺ + 2OH⁻,</em>

<em>Ba(OH)₂ is dissociated to Ba²⁺ and 2OH⁻.</em>

  • If H⁺ ions are added to the equilibrium:

H⁺ will combine with OH⁻ to form water.

<em>So, the concentration of OH⁻ will decrease and the equilibrium is disturbed.</em>

<em />

<em>According to Le Châtelier's principle: </em>when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.

  • So, the reaction will move to the right to suppress the effect of decreasing OH⁻ concentration.

  • The base will dissociate to form more OH⁻ and thus, the quantity of Ba(OH)₂ will decrease.

<em>So, the right choice is: the reaction will move to the left, is the choice that will not happen to the equilibrium.</em>

5 0
3 years ago
Someone please help me and answer my question asap. i need to turn my assignment in soon... please.
Ksenya-84 [330]

Answer:

Explanation:

a. the salt produced would be Mg3N2(magnesium nitride)

b. magnesium loses 2 electron to form Mg2+ ion and nitrogen gains 3 electron to form n3-

when several of these ions come together 3 Mg2+ ion combine with 2 n3- ion to form Mg3N2 thus Mg getting six electron from nitrogen to form a ionic bond.

c. the reaction is not balanced Mg + N2 = Mg3n2

to make it balanced the reaction should be 3 Mg + N2 = Mg3N2.

the reaction was not balanced before because the number of Mg on both side of the reaction was not equal.

d. magnesium nitrate has formula Mg(NO3)2 is formed when Mg combines with nitrogen and oxygen Mg + N2 + o2

8 0
3 years ago
PLEASE ASAP
Kitty [74]
Choice c - eats away other materials
7 0
3 years ago
Here are some data from a similar experiment, to determine the empirical formula of an oxide of tin. Calculate the empirical for
eduard

Answer:

Empirical formula of the Tin oxide sample is SnO₂

Explanation:

Tin reacts with combines with oxygen to form an oxide of tin.

Mass of crucible with cover = 19.66 g

Mass of crucible, cover, and tin sample = 22.29 g

Mass of crucible and cover and sample, after prolonged heating gives constant weight = 21.76 g

Mass of Tin oxide sample = 22.29 - 19.66 = 2.63 g

Mass of ordinary tin, after heating to breakdown the tin and oxygen = 21.76 - 19.66 = 2.1 g

Meaning that, mass of oxygen in the tin oxide sample = 2.63 - 2.1 = 0.53 g

Mass of Tin in the Tin Oxide sample = 2.1 g

Mass of Oxygen in the Tin oxide sample = 0.53 g

Convert these to number of moles

Number of moles of Tin on the Tin oxide sample = 2.1/118.71 = 0.0177

Number of moles of Oxygen in the Tin oxide sample = 0.53/16 = 0.0335

divide the number of moles by the lowest number

0.0177:0.0335

It becomes,

1:2

SnO₂

Hence, the empirical formula for the Tin oxide sample = SnO₂

7 0
4 years ago
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