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3 years ago

Key Questions and Terms Notes

Chemistry

1 answer:

Elanso [62]3 years ago

5 0

Answer:

Following are the solution to the given question:

Explanation:

Its interactions among light waves as well as different media were also called light products.
Even before sound rays hit a medium, the light rays bounce off.
Even at a similar angle with which the material initially struck its medium, sunlight reflecting.
When the light hits a mirror at such an angle of forty degrees, everything just rebonds at an angle of 40 degrees.
When light waves pass via a transparent medium, those who bend as well as change their direction.
Light is called transparent if it can pass via an object.
Whenever light waves strike a platform, this is neither re-elected nor refracted, nor expected to move.
The light of interacts with these materials are:

Reflection Mirror:

Water glass: refractive

Dark fabricate: Absorption

Sound waves and waves of water are examples of matter waves.
Light waves and X-rays are electromagnetic waves.
There is no need for sound radiation to pass through a medium/requirement to communicate with other particles to travel like waves of matter.
No, even though sound waves didn't reach through the space vacuum.
Matter waves should interact with solid, liquid, or gas particulates to travel.

The light waves pass thru the air more quickly than sound waves.

Sound waves almost always go through a metal frame.

Light waves travel faster than electrical signals even though they don't have to crash in electromagnetic waves to particles.

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4 years ago

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True or false. Gas particles never touch each other.

Sati [7]

Answer:

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Explanation:

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3 years ago

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: OCl−+I−→OI−+Cl− T

motikmotik

Answer :

(a) The rate law for the reaction is:

$\text{Rate}=k[OCl^-]^1[I^-]^1$

(b) The value of rate constant is, $60.4M^{-1}s^{-1}$

(c) rate of the reaction is $6.52\times 10^{-5}Ms^{-1}$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$OCl^-+I^-\rightarrow OI^-+Cl^-$

Rate law expression for the reaction:

$\text{Rate}=k[OCl^-]^a[I^-]^b$

where,

a = order with respect to $OCl^-$

b = order with respect to $I^-$

Expression for rate law for first observation:

$1.36\times 10^{-4}=k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b$ ....(1)

Expression for rate law for second observation:

$2.72\times 10^{-4}=k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b$ ....(2)

Expression for rate law for third observation:

$2.72\times 10^{-4}=k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b$ ....(3)

Dividing 1 from 2, we get:

$\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}\\\\2=2^a\\a=1$

Dividing 1 from 3, we get:

$\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b}\\\\2=2^b\\b=1$

Thus, the rate law becomes:

$\text{Rate}=k[OCl^-]^a[I^-]^b$

a = 1 and b = 1

$\text{Rate}=k[OCl^-]^1[I^-]^1$

Now, calculating the value of 'k' (rate constant) by using any expression.

$1.36\times 10^{-4}=k(1.5\times 10^{-3})(1.5\times 10^{-3})$

$k=60.4M^{-1}s^{-1}$

Now we have to calculate the rate for a reaction when concentration of $OCl^-$ and $I^-$ is $1.8\times 10^{-3}M$ and $6.0\times 10^{-4}M$ respectively.