The correct answer here is B - when chlorine accepts an electron to
complete its octet and becomes a chlorion ion it becomes an Anion. An
anion is a negatively charged ion. Chloride ions are an important
electrolyte within the body.
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:
There will remain 8.06 grams of ethane
Explanation:
Step 1: Data given
Mass of ethane = 9.32 grams
Mass of oxygen = 12.0 grams
Molar mass ethane = 30.07 g/mol
Molar mass oxygen = 32.00 g/mol
Step 2: The balanced equation
2 C2H6 + 7 O2 → 4 CO2 + 6 H2O
Step 3: Calculate moles ethane
Moles ethane = mass ethane / molar mass ethane
Moles ethane = 9.32 grams / 30.07 g/mol
Moles ethane = 0.3099 moles
Step 4: Calculate moles oxygen
Moles oxygen = 12.0 grams / 32.0 g/mol
Moles oxygen = 0.375 moles
Step 5: Calculate the limiting reactant
For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O
O2 is the limiting reactant. It will completely be consumed ( 0.375 moles)
Ethane is in excess. There will react 2/7 * 0.375 = 0.107 moles
There will remain 0.375 - 0.107 = 0.268 moles
Step 6: Calculate mass ethane
Mass ethane = moles ethane * molar mass ethane
Mass ethane = 0.268 moles * 30.07 g/mol
Mass ethane = 8.06 grams
There will remain 8.06 grams of ethane
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The true answer is: It's conserved because the total number of H atoms on each side is 12.
the first two answer is wrong because it's conserved not as mentioned, It's not conserved.
and the last one also wrong because the total number of O atoms are equal at the two sides but not equal 2.
