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SVEN [57.7K]
3 years ago
6

In the flame test, barium ions produce a green flame whereas calcium ions produce a red flame. In your own words, explain these

observations in terms of the electron structure of these two ions. In other words explain how these different colors are produced and why using your understanding of atomic structure.
Chemistry
2 answers:
FromTheMoon [43]3 years ago
5 0

Answer:

The different colours produced come from the different wavelength of the spectrum of light as the excited electrons in the metals return to their original state on cooling down.

Explanation:

The flame test is an analytical procedure used to identify different metals by passing them through a flame.

The flame, which is a source of heat energy heats up the metals and excites their outermost electrons which transit to the next energy level. On cooling down, as the excited electrons return to their original state, they emit different colours corresponding to the spectrum of light. Electrons of larger atoms like Barium emit light of higher frequency and lower wavelength and hence exhibits colours closer to the ultraviolet part of the electromagnetic spectrum while smaller atoms like Calcium emit light of lower frequency and higher wavelength and emit light closer to the infrared part of the spectrum

Baruim (atomic number 56) and Calcium (atomic number 20) are both metals in group two of the periodic table.

They both have two electrons in their outermost shell and are represented by Ba²⁺ and Ca²⁺.

When both metals are passed to through the flame test, the two outermost electrons in both metals are excited.

However because the outermost electrons in Barium are farther to the nucleus than those in Calcium, it takes a lesser amount of heat energy to excite them than that of Calcium which is closer to the nucleus.

The spectrum of light which is usually represented by 'ROYGBIV' has the following colours : Red, Orange,Yellow, Green, Indigo and Violet with Red having their highest wavelength (or shortest frequency) and Violet the shortest wavelength (or highest frequency).

Barium's green colour is because it outermost electrons emit light in the range of the Green spectrum of light and calcium's red colour is because its outermost electrons emit light in the range of the Red spectrum of light

deff fn [24]3 years ago
3 0

Answer:

The Barium flame is green because it is a difficult flame to excite, therefore for it to trigger a flame it is necessary that it be too excited for it to occur.

The reddish color of calcium is due to its high volatility and it is sometimes very difficult to differentiate it from strontium.the compression of these elements is due to being able to make them work during combustion

Explanation:

The flame test is a widely used qualitative analysis method to identify the presence of a certain chemical element in a sample. To carry it out you must have a gas burner. Usually a Bunsen burner, since the temperature of the flame is high enough to carry out the experience (a wick burner with an alcohol tank is not useful). The flame temperature of the Bunsen burner must first be adjusted until it is no longer yellowish and has a bluish hue to the body of the flame and a colorless envelope. Then the tip of a clean platinum or nichrome rod (an alloy of nickel and chromium), or failing that of glass, is impregnated with a small amount of the substance to be analyzed and, subsequently, the rod is introduced into the flame, trying to locate the tip in the least colored part of the flame.

The electrons in these will jump to higher levels from the lower levels and immediately (the time that an electron can be in higher levels is of the order of nanoseconds), they will emit energy in all directions in the form of electromagnetic radiation (light) of frequencies characteristics. This is what is called an atomic emission spectrum.

At a macroscopic level, it is observed that the sample, when heated in the flame, will provide a characteristic color to it. For example, if the tip of a rod is impregnated with a drop of Ca2 + solution (the previous notation indicates that it is the calcium ion, that is, the calcium atom that has lost two electrons), the color observed is brick red .

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Explanation :

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2Ag^{+}(aq)+Zn(s)\rightarrow 2Ag(s)+Zn^{2+}(aq)

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Q_c=\frac{[Zn^{2+}]}{[Ag^{+}]^2}

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I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r
maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P

Now, the moles of Li3P consumed by 15 g of Al2O3:

n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP

Therefore, the total mass of products is:

m_{products}=13g+8.5g\\\\m_{products}=21.5g

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

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3 years ago
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