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.
Answer:
4Fe + 3O2 + 6H2O → 4Fe(OH)3
Explanation:
The chemical formula for rust is Fe2O3 and is commonly known as ferric oxide or iron oxide. The final product is a series of chemical reactions simplified below as- The rusting of the iron formula is simply 4Fe + 3O2 + 6H2O → 4Fe(OH)3. The rusting process requires both the elements of oxygen and water.
Answer:
a) Mo the electron configuration: 42Mo: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d4
Mo3+ - is Paramagnetic
b) Au - [Xe] 4f14 5d10 6s1
For Au+ is not paramagnetic
c) Mn - [Ar] 3d5 4s2
Mn2+ is paramagnetic
d)Hf -[Xe] 4f¹⁴ 5d² 6s²
Hf2+ is not paramagnetic
Explanation:
An atom becomes positively charged when it looses an electron.
Diamagnetism in atom occurs whenever two electrons in an orbital paired equalises with a total spin of 0.
Paramagnetism in atom occurs whenever at least one orbital of an atom has a net spin of electron. That is a paramagnetic electron is just an unpaired electron in the atom.
Here is a twist even if an atom have ten diamagnetic electrons, the presence of at least one paramagnetic electron, makes it to be considered as a paramagnetic atom.
Simply put paramagnetic elements are one that have unpaired electrons, whereas diamagnetic elements do have paired electron.
In the question, we are told that there are;
- A loaf containing 33 slices
- A loaf containing 33 slices A package of cheese containing 15 slices
We also know that he is making a sandwich that has 2 pieces of both cheese and bread.
Hence;
Total number of bread and cheese = 33 + 15.
Each loaf should have two pieces of each bread and the cheeses make a total of four pieces.
Therefore he can make = 33 + 15/4 = 12 sandwiches.
So you need to find the volume in L? If so:
Convert the mass of Lithium Bromide into moles by dividing the 100 grams by the molar mass of LiBr, taken from the periodic table
In a solution, moles = (concentration in mole/L) x (volume in L)
We know the moles, we have the concentration in mole/L, now find the volume in L, and you should get 0.288. Plz do the math and check for yourself
