The two ways to measure the sand in a sand castle are by counting the number of pails of sand used to build the castle and by determining the mass of the sand used in building the castle. The first method makes use of the volume of the pail to determine the amount of sand while the second method is a more quantitative way.<span />
Answer:
The answer is B. A hydrogen atom forms a convalent bond.........
Answer: Option (3) is the correct answer.
Explanation:
Atomic number of lithium is 3 and its electronic distribution is 2, 1. So, to attain stability it will loose an electron and hence, it forms a single bond.
Atomic number of chlorine is 17 and it has 7 valence electrons. Hence, in order to attain stability it will gain one electron and therefore, it forms a single bond only.
Atomic number of nitrogen is 7 and its electronic distribution is 2, 5. Therefore, to attain stability it needs to gain 3 more electrons. Hence, a nitrogen atom is able to form a triple bond and also it is able to form a double bond.
Hydrogen has atomic number 1 and it attains stability by gaining one electron. Therefore, a hydrogen atoms always forms a single bond.
Atomic number of fluorine is 9 and its electronic distribution is 2, 7. To complete its octet it needs to gain one electron. Hence, a fluorine atom always forms a single bond.
Thus, we can conclude that out of the given options nitrogen is most likely to form multiple (double or triple) bonds.
Answer:
It decreases.
Explanation:
Due to changes in the Coulombic force, the protons within the nucleus of the ion have a much easier time pulling at one fewer electrons. This way, the electrons are pulled closer to the center, causing the radius to decrease.
Answer:
The oxidizing agent is the MnO₄⁻
Explanation:
This is the redox reaction:
10 I⁻ (aq) + 2 MnO₄⁻ (aq) + 16 H⁺ (aq) → 5 I₂ (s) + 2 Mn²⁺ (aq) + 8 H2O (l)
Let's determine the oxidation and the reduction.
I⁻ acts with -1 in oxidation state and changes to 0, at I₂.
All elements in ground state has 0 as oxidation state.
As the oxidation state has increased, this is the oxidation, so the iodide is the reducing agent.
In the permanganate (MnO₄⁻), Mn acts with +7 in oxidation state and decreased to Mn²⁺. As the oxidation state is lower, we talk about the reduction. Therefore, the permanganate is the oxidizing agent because it oxidizes iodide to iodine
