Seven diatomic elements are H₂, Cl₂, N₂, F₂, Br₂, I₂ and O₂.
<h3>Which are diatomic molecules?</h3>
Diatomic molecules are those molecules in which two atoms of same elements are present, and they are combined to attain the stability.
The seven diatomic molecules which are exist in the chemistry are:
- Hydrogen gas (H₂)
- Chlorine gas (Cl₂)
- Nitrogen gas (N₂)
- Fluorine gas (F₂)
- Bromine gas (Br₂)
- Iodine gas (I₂)
- Oxygen gas (O₂)
Hence H₂, Cl₂, N₂, F₂, Br₂, I₂ and O₂ are 7 diatomic molecules.
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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.
<u>We are given:</u>
Mass of ice = 21 grams
The ice is already at 0°c, the temperature at which it melts to form water
Molar heat of fusion of Ice = 6.02 kJ/mol
<u>Finding the energy required:</u>
<u>Number of moles of Ice: </u>
Molar mass of water = 18 g/mol
Number of moles = given mass/ molar mass
Number of moles = 21 / 18 = 7/6 moles
<u>Energy required to melt the given amount of ice:</u>
Energy = number of moles * molar heat of fusion
Energy = (7/6) * (6.02)
Energy = 7.02 kJ OR 7020 joules
Depending upon the clumping reaction with anti A , anti B and anti Rh antibodies the blood types are determined.
Explanation:
Agglutination (clumping) will occur when blood that contains the particular antigen is mixed with the particular antibody.
A+ have Agglutination with Anti-A ,Anti-Rh and No agglutination with Anti-B.
A- have Agglutination with Anti-A and No agglutination with Anti-B and Anti-Rh.
B+ have Agglutination with Anti-B Anti-Rh and No agglutination with Anti-A.
B- have Agglutination with Anti-B and No agglutination with Anti-B and Anti-Rh.
Rh+ have Agglutination with Anti-A and Anti-Rh and No agglutination with Anti-B.
Rh- have No Agglutination with Anti-A and Anti-B and Anti-Rh.
Endothermic reactions are reactions that require heat in the course of the process. The heat of reaction in this case is positive which means the energy of the products is greater than the energy of the reactants. In this case, the answer to this problem would have to be <span>1. must be greater than the required ΔH </span>
