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egoroff_w [7]
2 years ago
11

Why are coloures not observed in reflected light with light in the case of thick film?​

Chemistry
1 answer:
Dovator [93]2 years ago
7 0

Answer:

Explanation:

Thick plates doesnt show interference pattern as the optical path difference is greater than coherency length. Hence no fringe formation takes place and colours don't segregate and hece all light is seen as white light.

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How many alkali metals will combine with 1 oxygen family member
Daniel [21]
The alkali metals are so reactive that they are never found in nature in elemental form. Although some of their ores are abundant, isolating them from their ores is somewhat difficult. For these reasons, the group 1 elements were unknown until the early 19th century, when Sir Humphry Davy first prepared sodium (Na) and potassium (K) by passing an electric current through molten alkalis. (The ashes produced by the combustion of wood are largely composed of potassium and sodium carbonate.) Lithium (Li) was discovered 10 years later when the Swedish chemist Johan Arfwedson was studying the composition of a new Brazilian mineral. Cesium (Cs) and rubidium (Rb) were not discovered until the 1860s, when Robert Bunsen conducted a systematic search for new elements. Known to chemistry students as the inventor of the Bunsen burner, Bunsen’s spectroscopic studies of ores showed sky blue and deep red emission lines that he attributed to two new elements, Cs and Rb, respectively. Francium (Fr) is found in only trace amounts in nature, so our knowledge of its chemistry is limited. All the isotopes of Fr have very short half-lives, in contrast to the other elements in group 1.
7 0
3 years ago
Hydrogen iodide, HI, is formed in an equilibrium reaction when gaseous hydrogen and iodine gas are heated together. If 20.0 g of
Kaylis [27]

Answer: D. 19.9 g hydrogen remains.

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of H_2

\text{Number of moles}=\frac{20.0g}{2g/mol}=10.0moles

b) moles of I_2

\text{Number of moles}=\frac{20.0g}{254g/mol}=0.0787moles

H_2(g)+I_2(g)\rightarrow 2HI(g)

According to stoichiometry :

1 mole of I_2 require 1 mole of H_2

Thus 0.0787 moles of l_2 require=\frac{1}{1}\times 0.0787=0.0787moles of H_2

Thus l_2 is the limiting reagent as it limits the formation of product and H_2 acts as the excess reagent. (10.0-0.0787)= 9.92 moles of H_2are left unreacted.

Mass of H_2=moles\times {\text {Molar mass}}=9.92moles\times 2.01g/mol=19.9g

Thus 19.9 g of H_2 remains unreacted.

5 0
2 years ago
What purpose do each of the following play in the isolation and purification of ethyl 3- hydroxybutanoate?
Shalnov [3]

Explanation:

-Filter help — delete some big unreacted, undesirable species (norit is probably from what you are sorting through, its only carbon which cleans up things)

— extract with DCM because you are probably in an aqueous phase, and some butanoate is in it

- Anhydrous sodium absorbs excess of  water (dries the material)

-evaporation in the hood to clear the DCM and crystallize the product.

8 0
3 years ago
Atoms of which two elements have combined total of 23 protons​
vichka [17]

Answer:

sodium and magnesium

Explanation:

4 0
3 years ago
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A physiologist is studying the homeostatic control of blood ph. What type of receptor might be responsible for detecting changes
sertanlavr [38]

Answer:

Chemoreceptors should be your answer :)

Explanation:

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