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Zigmanuir [339]
3 years ago
12

What observation about light supported Einstein's theory?

Chemistry
1 answer:
Llana [10]3 years ago
5 0

In 1905 Albert Einstein had proposed a solution to the problem of observations made on the behaviour of light having characteristics of both wave and particle theory. From work of Plank on emission of light from hot bodies, Einstein suggested that light is composed of tiny particles called <span>photons, </span>and each photon has energy.

Light theory branches in to the physics of <span>quantum mechanics, </span>which was conceptualised in the twentieth century. Quantum mechanics deals with behaviour of nature on the atomic scale or smaller.

As a result of quantum mechanics, this gave the proof of the dual nature of light and therefore not a contradiction.

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Consider the following reaction between mercury(II) chloride and oxalate ion.
Alina [70]

<u>Answer:</u> The rate law of the reaction is \text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2

<u>Explanation:</u>

Rate law 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:

2 HgCl_2(aq.)+C_2O_4^{2-}(aq.)\rightarrow 2Cl^-(aq.)+2CO_2(g)+Hg_2Cl_2(s)

Rate law expression for the reaction:

\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b

where,

a = order with respect to HgCl_2

b = order with respect to C_2O_4^{2-}

Expression for rate law for first observation:

3.2\times 10^{-5}=k(0.164)^a(0.15)^b  ....(1)

Expression for rate law for second observation:

2.9\times 10^{-4}=k(0.164)^a(0.45)^b  ....(2)

Expression for rate law for third observation:

1.4\times 10^{-4}=k(0.082)^a(0.45)^b  ....(3)

Expression for rate law for fourth observation:

4.8\times 10^{-5}=k(0.246)^a(0.15)^b  ....(4)  

Dividing 2 from 1, we get:

\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{(0.164)^a(0.45)^b}{(0.164)^a(0.15)^b}\\\\9=3^b\\b=2

Dividing 2 from 3, we get:

\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{(0.164)^a(0.45)^b}{(0.082)^a(0.45)^b}\\\\2=2^a\\a=1

Thus, the rate law becomes:

\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2

3 0
3 years ago
These models show the electron structures of two different nonmetal elements. Element 1 at left has a purple circle at center wi
Daniel [21]

Answer:

Element 2

Explanation:

If we look at the model stated for element 1, it is clear that element 1 must be a noble gas. It has eight electrons in its outermost shell this implies that it has already attained a complete octet of electrons and is reluctant towards chemical reaction.

The second element belongs to group 16 since it has six electrons on its outermost shell. It is certainly more reactive than element 1 which is a noble gas.

7 0
3 years ago
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Identify each of the atomic models described here. Atoms are indivisible spheres. plum pudding model Dalton model Bohr model
mamaluj [8]

Atoms are indivisible spheres-Dalton model

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What are the statements?.
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Sulfur dioxide gas (SO2) reacts with excess oxygen gas (O2) and excess liquid water (H2O) to form liquid sulfuric acid (H2SO4).
inessss [21]
A.2SO₂ + O₂ + 2H₂O → 2H₂SO₄
B.Moles of SO₂ = 67.2 / 22.4Moles of SO₂ = 3 molesMoles of H₂SO₄ = 3 molesMass of H₂SO₄ = 3 x 98Mass of H₂SO₄ = 294 grams
Assuming sulfuric acid to have the same density as water,density = 1000 g / LVolume = mass / densityVolume = 294 / 1000Volume = 0.29 liters of sulfuric acid

6 0
3 years ago
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