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erik [133]
2 years ago
8

What allowed Mendeleev to make predictions of undiscovered elements

Chemistry
2 answers:
Mademuasel [1]2 years ago
7 0
Predictions using gaps Mendeleev left gaps in his table to place elements not known at the time. By looking at the chemical properties and physical properties of the elements next to a gap, he could also predict the properties of these undiscovered elements.
Andrej [43]2 years ago
3 0

Answer:

he predicted the properties from known elements above and belws the unknown in the same group

Explanation:

What allowed Mendeleev to make predictions of undiscovered elements

He realized that an element on this table with one known element above it and one known  element below it had to have properties between the two known elements

How did Mendeleev predict gallium and germanium?

Based on gaps in the periodic table Mendeleev deduced that in these gaps belonged elements yet to be discovered. Based on other elements below and above in the same group he predicted the existence of eka-aluminum, eka-boron, and eka-silicon, later to be named gallium (Ga), scandium (Sc), and germanium (Ge).

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In each of the following sets of elements, which one will be least likely to gain or lose electrons?
klasskru [66]
1. The reactivity among the alkali metals increases as you go down the group due to the decrease in the effective nuclear charge from the increased shielding by the greater number of electrons. The greater the atomic number, the weaker the hold on the valence electron the nucleus has, and the more easily the element can lose the electron. Conversely, the lower the atomic number, the greater pull the nucleus has on the valence electron, and the less readily would the element be able to lose the electron (relatively speaking). Thus, in the first set comprising group I elements, sodium (Na) would be the least likely to lose its valence electron (and, for that matter, its core electrons).

2. The elements in this set are the group II alkaline earth metals, and they follow the same trend as the alkali metals. Of the elements here, beryllium (Be) would have the highest effective nuclear charge, and so it would be the least likely to lose its valence electrons. In fact, beryllium has a tendency not to lose (or gain) electrons, i.e., ionize, at all; it is unique among its congeners in that it tends to form covalent bonds.

3. While the alkali and alkaline earth metals would lose electrons to attain a noble gas configuration, the group VIIA halogens, as we have here, would need to gain a valence electron for an full octet. The trends in the group I and II elements are turned on their head for the halogens: The smaller the atomic number, the less shielding, and so the greater the pull by the nucleus to gain a valence electron. And as the atomic number increases (such as when you go down the group), the more shielding there is, the weaker the effective nuclear charge, and the lesser the tendency to gain a valence electron. Bromine (Br) has the largest atomic number among the halogens in this set, so an electron would feel the smallest pull from a bromine atom; bromine would thus be the least likely here to gain a valence electron.

4. The pattern for the elements in this set (the group VI chalcogens) generally follows that of the halogens. The greater the atomic number, the weaker the pull of the nucleus, and so the lesser the tendency to gain electrons. Tellurium (Te) has the highest atomic number among the elements in the set, and so it would be the least likely to gain electrons.
7 0
2 years ago
a compound has a molar mass of 129 g/mol if its empirical formula is C2H5N then what is the molecular formula
Elena-2011 [213]

Given :

A compound has a molar mass of 129 g/mol .

Empirical formula of compound is C₂H₅N .

To Find :

The molecular formula of the compound.

Solution :

Empirical mass of compound :

M_e = ( 2 \times 12 ) + ( 5 \times 1 ) + (  1  \times 14 )\\\\M_e = 43\ gram/mol

Now, n-factor is :

n = \dfrac{M}{M_e}\\\\n = \dfrac{129}{43}\\\\n = 3

Multiplying each atom in the formula by 3 , we get :

Molecular Formula, C₆H₁₅N₃

3 0
2 years ago
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