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.
The concentration of [H3O⁺]=2.86 x 10⁻⁶ M
<h3>Further explanation</h3>
In general, the weak acid ionization reaction
HA (aq) ---> H⁺ (aq) + A⁻ (aq)
Ka's value
![\large {\boxed {\bold {Ka \: = \: \frac {[H ^ +] [A ^ -]} {[HA]}}}}](https://tex.z-dn.net/?f=%5Clarge%20%7B%5Cboxed%20%7B%5Cbold%20%7BKa%20%5C%3A%20%3D%20%5C%3A%20%5Cfrac%20%7B%5BH%20%5E%20%2B%5D%20%5BA%20%5E%20-%5D%7D%20%7B%5BHA%5D%7D%7D%7D%7D)
Reaction
HC₂H₃O₂ (aq) + H₂O (l) ⇔ (aq) + H₃O⁺ (aq) Ka = 1.8 x 10⁻⁵
![\tt Ka=\dfrac{[C_2H_3O^{2-}[H_3O^+]]}{[HC_2H_3O_2]}}\\\\1.8\times 10^{-5}=\dfrac{0.22\times [H_3O^+]}{0.035}](https://tex.z-dn.net/?f=%5Ctt%20Ka%3D%5Cdfrac%7B%5BC_2H_3O%5E%7B2-%7D%5BH_3O%5E%2B%5D%5D%7D%7B%5BHC_2H_3O_2%5D%7D%7D%5C%5C%5C%5C1.8%5Ctimes%2010%5E%7B-5%7D%3D%5Cdfrac%7B0.22%5Ctimes%20%5BH_3O%5E%2B%5D%7D%7B0.035%7D)
[H₃O⁺]=2.86 x 10⁻⁶ M
Answer:
116 g
Explanation:
From the question given above, the following data were obtained:
Number of mole of calcium = 2.9 moles
Mass of calcium =.?
The mole and mass of a substance are related according to the following formula:
Mole = mass / molar mass
With the above formula, we can obtain the mass of calcium. This can be obtained as follow:
Number of mole of calcium = 2.9 moles
Molar mass of calcium = 40 g/mol
Mass of calcium =.?
Mole = mass / molar mass
2.9 = mass of calcium / 40
Cross multiply
Mass of calcium = 2.9 × 40
Mass of calcium = 116 g
Therefore, the mass of 2.9 moles of calcium is 116 g.
Answer:
The answer to your question is: letter D.
Explanation:
Noble gases are located in group VIIIA of the periodic table, this means that they have 8 eight electrons in their outermost shell.
Due to this characteristic, they are stable and do not react with other elements.
a. 1s22s22p4 The outermost shell of this electron configuration has 6 electrons, then this element has 6 electrons not 8. This configuration is of an element of the group VIA.
b. [Ne]2s22p2 The outermost shell of this element has 4 electrons, so this is not the configuration of a noble gas.
c. [Ar] 3s1 This element only has one electron in its outermost shell, so this is the electron configuration of an alkaline metal.
d. 1s22s22p6 This element has 8 electrons in its outermost shell, so this is the electron configuration of a noble gas.
When diluting solutions from concentrated solutions the following formula can be used
c1v1 = c2v2
where c1 is concentration and v1 is volume of the concentrated solution
and c2 is concentration and v2 is volume of the diluted solution to be prepared
substituting these values
0.0813 M x 16.5 mL = 0.0200 M x V
V = 67.1 mL
the volume of the diluted solution prepared is 67.1 mL.
the volume of water that should be added to get a final volume of 67.1 mL is (67.1 - 16.5 ) = 50.6 mL
a volume fo 50.6 mL should be added
