Answer:
B. the resonance hybrid of all structures
Explanation:
The idea of resonance is used to explain bonding in compounds where a single structure does not fully account for all the bonding interactions in a molecule.
A number of equivalent structures are then used to show the nature of bonding in such a molecule. Such structures are called resonance structures or canonical structures. None of these structures individually offer a holistic explanation to the bonding interactions in the molecule under study.
However, a hybrid of all the canonical structures does explain the nature of bonding in the molecule.
Answer:
MEANS:
1 = 97.7
2 = 74.3
3 = 50
4 = 30
5 = 13
UNCERTAINTY:
gimme a sec, ill put it in the comments under this
Explanation:
Brittleness is a characteristic that describes chalk. Its color and shape also describe the chalk. Any such characteristic of a material that you can observe without changing the substances that make up the material is a physical property
Gold is metallic, with a yellow colour when in a mass, but when finely divided it may be black, ruby, or purple. It is the most malleable and ductile metal
Answer:
See detailed answer with explanation below.
Explanation:
Valence electrons are electrons found on the outermost shell of an atom. They are the electrons in an atom that participate in chemical combination. Recall that the outermost shell of an atom is also referred to as its valence shell. Let us consider an example; if we look at the atom, sodium-11, its electronic configuration is 2,8,1. The last one electron is the valence electron of sodium which is found in its outermost or valence shell.
Positive ions are formed when electrons are lost from the valence shell of an atom. For instance, if the outermost electron in sodium is lost, we now form the sodium ion Na^+ which is a positive ion. Positive ions possess less number of electrons compared to their corresponding atoms.
Negative ions are formed when one or more electrons is added to the valence shell of an atom. A negative ion possesses more electrons than its corresponding atom. For example, chlorine(Cl) contains 17 electrons but the chloride ion (Cl^-) contains 18 electrons.
In molecular compounds, a bond is formed when two electrons are shared between the bonding atoms. Each bonding atom may contribute one of the shared electrons (ordinary covalent bond) or one of the bonding atoms may provide the both shared electrons (coordinate covalent bond). The shared pair may be located at an equidistant position to the nucleus of both atoms. Similarly, the electron may be drawn closer to the nucleus of one atom than the other (polar covalent bond) depending on the electro negativity of the two bonding atoms.
The electrons are shared in order to complete the octet of each atom by so doing, the both bonding atoms now obey the octet rule. For example, two chlorine atoms may come together to form a covalent bond in which each chlorine atom has an octet of electrons on its outermost shell.
1) Calculate the number of moles of O2 (g) in 300 cm^3 of gas at 298 k and 1 atm
Ideal gas equation: pV = nRT => n = pV / RT
R = 0.0821 atm*liter/K*mol
V = 300 cm^3 = 0.300 liter
T = 298 K
p = 1 atm
=> n = 1 atm * 0.300 liter / [ (0.0821 atm*liter /K*mol) * 298K] = 0.01226 mol
2) The reaction of a metal with O2(g) to form an ionic compound (with O2- ions) is of the type
X (+) + O2 (g) ---> X2O or
2 X(2+) + O2(g) ----> X2O2 = 2XO or
4X(3+) + 3O2(g) ---> 2X2O3
In the first case, 1 mol of metal react with 1 mol of O2(g); in the second case, 2 moles of metal react with 1 mol of O2(g); in the third, 4 moles of X react with 3 moles of O2(g)
So, lets probe those 3 cases.
3) Case 1: 1 mol of metal X / 1 mol O2(g) = x moles / 0.01226 mol
=> x = 0.01226 moles of metal X
Now you can calculate the atomic mass of the hypotethical metal:
1.15 grams / 0.01226 mol = 93.8 g / mol
That does not correspond to any of the metal with valence 1+
So, now probe the case 2.
4) Case 2:
2moles X metal / 1 mol O2(g) = x / 0.01226 mol
=> x = 2 * 0.01226 = 0.02452 mol
And the atomic mass of the metal is: 1.15 g / 0.02452 mol = 46.9 g/mol
That is similar to the atomic mass of titanium which is 47.9 g / mol and whose valece is 2+.
4) Case 3
4 mol meta X / 3 mol O2 = x / 0.01226 => x = 0.01226 * 4 / 3 = 0.01635
atomic mass = 1.15 g / 0.01635 mol = 70.33 g/mol
That does not correspond to any metal.
Conclusion: the identity of the metallic element could be titanium.
