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Brrunno [24]
3 years ago
13

How did Henry Moseley revise Mendeleev's periodic table? (Please answer A.S.A.P.)

Chemistry
2 answers:
CaHeK987 [17]3 years ago
5 0
<span>If you give it a good search, the most used answer would probably be as follows,

</span><span>In 1914 Henry Moseley found a relationship between an element's X-ray wavelength and its atomic number (Z), and therefore rearranged the table by nuclear charge / atomic number rather than atomic weight. Before this discovery, atomic numbers were just sequential numbers based on an element's atomic weight. Moseley's discovery showed that atomic numbers had an experimentally measurable basis.
</span>
Hope this helps!
jekas [21]3 years ago
5 0

Answer:

Arranged the elements on the table in order of increasing number of protons, or atomic number.

Explanation:

Mendeleev listed the known chemical elements in ascending order of atomic mass. However, such classification gave some problems to Mendeleev's table, which was characterized by the impression that some elements appeared to be out of place. An example was argon which, when isolated, did not appear to have the correct mass to justify its position. Its relative atomic mass of 40 is the same as that of calcium, but these differed considerably: while argon is an inert gas, calcium is a very reactive metal.

In the early twentieth century, when Henry Moseley examined the x-ray spectrum of the elements, he found that all atoms of the same chemical element had the same nuclear charge, and therefore had the same number of protons, which consist of the atomic number of the elements. It was quickly concluded that the elements would be in an even more regular pattern when arranged in a table in ascending order of their atomic number rather than atomic mass.

Thus, we can conclude that Moseley revised Mendeleev's periodic table by arranging the elements on the periodic table in ascending order of proton number or atomic number.

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4 years ago
determine the ph of a buffer that is 0.55 M HNO2 and 0.75 M KNO2. tha value of Ka for HNO2 is 6.8*10^-4
Mariana [72]

Answer:

pH = 3.3

Explanation:

Buffer solutions minimize changes in pH when quantities of acid or base are added into the mix. The typical buffer composition is a weak electrolyte (wk acid or weak base) plus the salt of the weak electrolyte. On addition of acid or base to the buffer solution, the solution chemistry functions to remove the acid or base by reacting with the components of the buffer to shift the equilibrium of the weak electrolyte left or right to remove the excess hydronium ions or hydroxide ions is a way that results in very little change in pH of the system. One should note that buffer solutions do not prevent changes in pH but minimize changes in pH. If enough acid or base is added the buffer chemistry can be destroyed.

In this problem, the weak electrolyte is HNO₂(aq) and the salt is KNO₂(aq). In equation, the buffer solution is 0.55M HNO₂ ⇄ H⁺ + 0.75M KNO₂⁻ . The potassium ion is a spectator ion and does not enter into determination of the pH of the solution. The object is to determine the hydronium ion concentration (H⁺) and apply to the expression pH = -log[H⁺].

Solution using the I.C.E. table:

              HNO₂ ⇄    H⁺   +   KNO₂⁻

C(i)        0.55M       0M      0.75M

ΔC            -x            +x          +x

C(eq)  0.55M - x       x     0.75M + x    b/c [HNO₂] / Ka > 100, the x can be                                    

                                                             dropped giving ...

           ≅0.55M        x       ≅0.75M        

Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]

=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M

pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3

Solution using the Henderson-Hasselbalch Equation:

pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]

= -log(6.8 x 10⁻⁴) + log[(0.75M)/(0.55M)]

= -(-3.17) + 0.14 = 3.17 + 0.14 = 3.31 ≅ 3.3

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Answer: All of these statements are true

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Melting point help us to determine if a mixture is pure or has impurities by the virtues of it melting range..

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