Boron’s chemistry is not typical of its group. is group 3A (13) shows the increasing metallic character from Al to Tl.
All Boron compounds are covalent whereas the other elements in group 3A (13) form mostly ionic compounds.
Except for Boron, the other elements of group 3A (13) show increasing metallic character from Al to Tl. But Boron is a metalloid.
Compared to the other elements in group 3A, boron has a lower reactivity in chemical terms (13)
The metalloid boron (B), as well as the metals aluminium (Al), gallium (Ga), indium (In), and thallium, are all part of group 3A (or IIIA) of the periodic table (Tl). In contrast to the other members of Group 3A, the element borax primarily forms covalent connections.
To learn more about group 3A (13) refer the link:
brainly.com/question/5489194
#SPJ4
Answer : The value of equilibrium constant for this reaction at 328.0 K is 
Explanation :
As we know that,
where,
= standard Gibbs free energy = ?
= standard enthalpy = 151.2 kJ = 151200 J
= standard entropy = 169.4 J/K
T = temperature of reaction = 328.0 K
Now put all the given values in the above formula, we get:
The relation between the equilibrium constant and standard Gibbs free energy is:
where,
= standard Gibbs free energy = 95636.8 J
R = gas constant = 8.314 J/K.mol
T = temperature = 328.0 K
K = equilibrium constant = ?
Now put all the given values in the above formula, we get:
Therefore, the value of equilibrium constant for this reaction at 328.0 K is
Can you restate the questions again ?
Volume well in the picture the fish tank seems to be a rectangle, so you would do V=length x width x height. So V=5x3x3=45
<span>The answer is 4. The molecules of each material entice each other over dispersion (London) intermolecular forces. Whether a substance is a solid, liquid, or gas hinge on the stability between the kinetic energies of the molecules and their intermolecular magnetisms. In fluorine, the electrons are firmly apprehended to the nuclei. The electrons have slight accidental to stroll to one side of the molecule, so the London dispersion powers are comparatively weak. As we go from fluorine to iodine, the electrons are far from the nuclei so the electron exhausts can more effortlessly misrepresent. The London dispersion forces developed to be increasingly stronger.</span>
