Any method of determining whether an event or object is older or younger than others

consider an atom having four distinct neergy levels. if an electron is able to make transitions between, any two levels, how man

ludmilkaskok [199]

Answer:

6 different frequencies

Explanation:

From energy level 1 to 2 is one frequency, from energy level 1 to 3 is one frequency and From energy level 1 to 4 is one frequency. So, we have a total of 3 frequencies for transition from energy level 1.

From energy level 2 to 3 is one frequency and from energy level 2 to 4 is one frequency. So, we have a total of 2 frequencies for transition from energy level 2.

From energy level 3 to 4 is one frequency.

So we have a total of 3 + 2 + 1 different frequencies = 6 different frequencies.

Note that the reverse process for each step produces the same frequency as the step in consideration.

8 0

2 years ago

Aluminium +hydrochloride acid ________+_________

REY [17]

Answer:

Aluminium Chloride + Hydrogen

Please vote for Brainliest and I hope this helps!

4 0

1 year ago

What substances pass through a leaf's stomata?

Lera25 [3.4K]

Oxygen and carbon dioxide

7 0

2 years ago

Read 2 more answers

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

2 years ago

Effect of Two-Step Homogenization on the Evolution of Al3Zr Dispersoids in Al-0.3Mg-0.4Si-0.2Zr Alloy

AlekseyPX

Effect of Two-Step Homogenization on the Evolution of Al3Zr Dispersoids in Al-0.3Mg-0.4Si-0.2Zr Alloy Al3Zr nano-particles can be introduced in Al-Mg-Si 6xxx alloys to improve their elevated temperature behavior and recrystallization resistance. The effect of two-step homogenization treatments on

the precipitation of Al3Zr dispersoids in Al-0.3Mg-0.4Si-0.2Zr alloy was investigated and compared to

<h3>What is Homogenization?</h3>

Any of a number of methods, including homogenization and homogenisation, are used to uniformly combine two liquids that are insoluble in one another. To do this, one of the liquids is changed into a state in which very minute particles are evenly dispersed across the other liquid. The process of homogenizing milk, in which the milk fat globules are equally distributed throughout the remaining milk and reduced in size, is a classic example. In order to create an emulsion, two immiscible liquids (i.e., liquids that are not soluble in all amounts one in another) must be homogenized (from "homogeneous"; Greek, homos, same + genos, kind)[2] (Mixture of two or more liquids that are generally immiscible).

To learn more about Homogenization from the given link:

brainly.com/question/18271118

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7 0

10 months ago