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sergeinik [125]
3 years ago
13

Increasing the concentration of a reactant shifts the position of a chemical equilibrium towards formation of more products. wha

t effect does adding a reactant have on the rates of the forward and reverse reactions?
a)There is no effect
b)Both the forward and reverse speed up by same amount
c)The foward reaction speeds up immeditaly
d)Forward reaction slows down
e)the forward reaction speeds up
f)more info needs to be given.
Chemistry
1 answer:
Natalija [7]3 years ago
4 0
<h2>The required "option is e)".</h2>

Explanation:

  • The Rate of formation of products depends on the concentration of reactants or the forward reaction increases on increasing the rate of the concentration.
  • There is no effect on the equilibrium rate when the concentration of reactants and products is constant.
  • Forward reaction slows down when the reactant concentration is decreased.
  • On increasing the amount or concentration of reactant the chemical  equilibrium shifts towards formation of more products.
  • Hence, the forward reaction speeds up.
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the atomic number of barium is 56. What do you know about the subatomic particles in an atom of this element ?
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4 years ago
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cupoosta [38]
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6 0
4 years ago
What element produces violet light of approximately 412 nm?
Natali [406]
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8 0
3 years ago
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An analytical chemist is titrating of a solution of hydrazoic acid with a solution of . The of hydrazoic acid is . Calculate the
serg [7]

Answer:

pH = 12.43

Explanation:

<em>...is titrating 212.7 mL of a 0.6800 M solution of hydrazoic acid (HN3) with a 0.2900 M solution of KOH. The p Ka of hydrazoic acid is 4.72. Calculate the pH of the acid solution after the chemist has added 571.6 mL of the KOH solution to it</em>.

To solve this question we need to know that hidrazoic acid reacts with KOH as follows:

HN3 + KOH → KN3 + H2O

<em>Moles KOH:</em>

0.5716L * (0.2900mol /L) =0.1658 moles of KOH

<em>Moles HN3:</em>

0.2127L * (0.6800mol/L) = 0.1446 moles HN3

As the reaction is 1:1, the KOH is in excess. The moles in excess of KOH are:

0.1658 moles - 0.1446 moles =

0.0212 mol KOH

In 212.7mL + 571.6mL = 784.3mL = 0.7843L

The molarity of KOH = [OH-] is:

0.0212 mol KOH / 0.7843L = 0.027M = [OH-]

The pOH is defined as -log [OH-]

pOH = -log 0.027M

pOH = 1.57

pH = 14 - pOH

pH = 12.43

6 0
3 years ago
Polymerization is the process of linking smaller molecules to form long chains of higher molecular weight. True or False
quester [9]
True is correct answer.


Polymerization is the process of linking it has a smaller molecules to form has a long chains of higher molecular weight.

Hope it helped you.

-Charlie
6 0
4 years ago
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