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2 years ago

What is the application of the emission spectra regarding elements?

1 answer:

8 0

The emission spectrum of a chemical element or chemical compound is the spectrum of frequencies of electromagnetic radiation emitted due to an atom or molecule making a transition from a high energy state to a lower energy state. The photon energy of the emitted photon is equal to the energy difference between the two states. There are many possible electron transitions for each atom, and each transition has a specific energy difference. This collection of different transitions, leading to different radiated wavelengths, make up an emission spectrum. Each element's emission spectrum is unique. Therefore, spectroscopy can be used to identify elements in matter of unknown composition. Similarly, the emission spectra of molecules can be used in chemical analysis of substances.

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If you only wanted to increase the particle motion of a gas without increasing any of its other properties, which would the most

andrey2020 [161]

Answer:

Explanation:

5 0

2 years ago

Hello,

nikitadnepr [17]

Answer:

$a)\ 2FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3\\b)\ Decomposition\ Reaction$

Explanation:

Ferrous Sulphate $(FeSO4)$ is generally found as Lime-Green Crystals. On heating, these crystals almost immediately turn white-yellow. They then, break down to produce an anhydrous mixture of Sulphur Trioxide $(SO_3)$ , Sulphur Dioxide $(SO_2)$ as well as Ferric Oxide $(Fe_2O_3)$ .

We can hence, frame a skeletal equation of this reaction and try to balance it.

Hence,

$FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3$

Now,

a)In order to balance it through the 'Hit &Trial Method', we'll follow a series of steps:

1. First, lets compare the number of Fe (Iron) atoms on the RHS and LHS. We find that, the no. of Fe Atoms on the RHS is twice the number of Fe Atoms on the LHS. We hence, add a co-effecient 2 beside $FeSO_4$ .

2. Now, Iron atoms, Sulphur Atoms and Oxygen atoms occur 2, 2, 8 respectively on both the sides:

Hence, As all the other elements as well as iron, balance, we've arrived upon our Balanced Equation :

$2FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3$

b) We know that, decomposition reactions are [generally] endothermic reactions in which Large Compounds decompose into smaller elements and compounds. Here, as Ferrous Sulphate decomposes into Sulphur Dioxide, Sulphur Trioxide and Ferric Oxide, the reaction that occurs here is Decomposition Reaction.

7 0

2 years ago

In the following reaction, what element is gaining mass?

vagabundo [1.1K]

Mg gained mass because it went from being a single element (on the reactant side) to being a molecule (on the product side).

8 0

2 years ago

The rate constant for the reaction 3A equals 4b is 6.00×10 how long will it take the concentration of a to drop from 0.75 to 0.2

Lelu [443]

This question is incomplete, the complete question is;

The rate constant for the reaction 3A equals 4B is 6.00 × 10⁻³ L.mol⁻¹min⁻¹.

how long will it take the concentration of A to drop from 0.75 to 0.25M ?

from the unit of the rate constant we know it is a second reaction order

OPTIONS

a) 2.2×10^−3 min

b) 5.5×10^−3 min

c) 180 min

d) 440 min

e) 5.0×10^2 min

Answer:

it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

Explanation:

Given that;

Rate constant K = 6.00 × 10⁻³ L.mol⁻¹min⁻¹

3A → 4B

given that it is a second reaction order;

k = 1/t [ 1/A - 1/A₀]

kt = [ 1/A - 1/A₀]

t = [ 1/A - 1/A₀] / k

K is the rate constant(6.00 × 10⁻³)

A₀ is initial concentration( 0.75 )

A is final concentration(0.25)

t is time required = ?

so we substitute our values into the equation

t = [ (1/0.25) - (1/0.75)] / (6.00 × 10⁻³)

t = 2.6666 / (6.00 × 10⁻³)

t = 444.34 ≈ 440 min {significant figures}

Therefore it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

6 0

2 years ago

Pls help asap!!!!

AVprozaik [17]

Answer: The pressure of the He is 2.97 atm

Explanation:

According to Dalton's law, the total pressure is the sum of individual pressures.

$p_{total}=p_{N_2}+p_{O_2}+p_{He}.$

Given : $p_{total}$ =total pressure of gases = 6.50 atm

$p_{N_2}$ = partial pressure of Nitrogen = 1.23 atm

$p_{O_2}$ = partial pressure of oxygen = 2.3 atm

$p_{He}$ = partial pressure of Helium = ?

putting in the values we get:

$6.50atm=1.23atm+2.3atm+p_{He}$

$p_{He}=2.97atm$

The pressure of the He is 2.97 atm

3 0

2 years ago