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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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5 0
3 years ago
What is the difference between solvation and dissociation
mel-nik [20]

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3 years ago
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Stels [109]
<h3>Further explanation</h3>

1.Atomic Number (Z) = Mass Number (A) - Number of Neutrons

neutrons = mass number-atomic number

Atomic mass Cl-37= 17

Mass number Cl-37=37

Neutrons = 37-17=20

2. Mass atom X = mass isotope 1 . % + mass isotope 2.%...

\tt atomic~mass~X=0.0224\times 88+0.7832\times 90+0.1944\times 91=90.1

3. The energy in one photon can be formulated as

\large{\boxed{\bold{E\:=\:h\:.\:f}}}

f = c / λ, so :

\tt E\approx f\approx \dfrac{1}{\lambda}

Energy is directly proportional to frequency and inversely proportional to the wavelength

So, as the frequency of photon increases, the energy of photon increases

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7 0
3 years ago
When trying to solve a social problem science does not take into account
VikaD [51]
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7 0
3 years ago
The midpoints of a regular hexagon are connected to form a smaller hexagon. The small hexagon has perimeter $2\sqrt{3}.$ What is
Murrr4er [49]
<h2>The perimeter of the large hexagon is equal to 12 units.</h2>

Explanation:

Given,

The perimeter of small hexagon = 2\sqrt{3}

To find, the perimeter of the large hexagon = ?

We know that,

The perimeter of the large hexagon

= \dfrac{6}{\sqrt{3}} × The perimeter of small hexagon

= \dfrac{6}{\sqrt{3}} \times 2\sqrt{3}

= 6 × 2

= 12 units

∴ The perimeter of the large hexagon = 12 units

Thus, the perimeter of the large hexagon is equal to 12 units.

4 0
4 years ago
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