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erastovalidia [21]
3 years ago
8

How can the rate of a reaction be increased?

Chemistry
2 answers:
Nana76 [90]3 years ago
4 0
I think it may be A.
But I'm not sure






Basile [38]3 years ago
4 0
The correct answer is A.

Hope this helps
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Describe the evidence of changes in organisms in the Cambrian Era that are reflected in the fossil record.​
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Answer:

Morphology and phylogenetics revealed by fossils. Perhaps the strongest evidence to support the Cambrian evolutionary explosion of animal forms is the first clear appearance, in the Early Cambrian, of skeletal fossils representing members of many marine bilaterian animal phyla

Explanation:

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3 years ago
In the reaction MgCl2 + 2KOH Mg(OH)2 + 2KCl, if 3 moles MgCl2 are added to 4 moles KOH, what determines how much Mg(OH)2 is made
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The correct option is C. The amount of MgCl2. we know this because <span>no matter how much you increase KOH, if you dont increase Mgcl2, the amount of Mg(OH)2 remains the same. Hope this works for you</span>
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3 years ago
N204(0) + 2NO2(g)
user100 [1]

setup 1 : to the right

setup 2 : equilibrium

setup 3 : to the left

<h3>Further explanation</h3>

The reaction quotient (Q) : determine a reaction has reached equilibrium

For reaction :

aA+bB⇔cC+dD

\tt Q=\dfrac{C]^c[D]^d}{[A]^a[B]^b}

Comparing Q with K( the equilibrium constant) :

K is the product of ions in an equilibrium saturated state  

Q is the product of the ion ions from the reacting substance  

Q <K = solution has not occurred precipitation, the ratio of the products to reactants is less than the ratio at equilibrium. The reaction moved to the right (products)

Q = Ksp = saturated solution, exactly the precipitate will occur, the system at equilibrium

Q> K = sediment solution, the ratio of the products to reactants is greater than the ratio at equilibrium. The reaction moved to the left (reactants)

Keq = 6.16 x 10⁻³

Q for reaction N₂O₄(0) ⇒ 2NO₂(g)

\tt Q=\dfrac{[NO_2]^2}{[N_2O_4]}

Setup 1 :

\tt Q=\dfrac{0.0064^2}{0.098}=0.000418=4.18\times 10^{-4}

Q<K⇒The reaction moved to the right (products)

Setup 2 :

\tt Q=\dfrac{0.0304^2}{0.15}=0.00616=6.16\times 10^{-3}

Q=K⇒the system at equilibrium

Setup 3 :

\tt Q=\dfrac{0.230^2}{0.420}=0.126

Q>K⇒The reaction moved to the left (reactants)

8 0
2 years ago
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