Answer: The correct choice is (b).
Explanation:
We have the mass of Magnesium at 97.22 g, and the molar mass of Magnesium is 24.305 g/mol.
Therefore, the calculation for the number of moles is as follows.
Number of moles = 
= 
= 4 mol
Additionally, it is known that one mole contains 
atoms/mol. Thus, we calculate the total number of atoms in 4 moles as follows.
= 
atoms
or, = 
atoms
Hence, we conclude that in 97.22 grams of Magnesium, there are atoms.
Answer:
Explanation:
In KCl, the two elements that combine to create KCl are potassium (K) and chlorine (Cl).
Potassium, as a Group 1 element, possesses one valence electron in its outermost shell which it readily donates during bonding. Every element aims to achieve a stable electron configuration, typically with 2 or 8 electrons in its outer shell. Potassium is characterized by its lower electronegativity and higher ionization energy, making it more likely to donate its electron than to accept one. On the other hand, chlorine belongs to Group 17 and has 7 electrons in its outer shell, requiring just one additional electron to complete its octet. Chlorine’s higher electronegativity and lower ionization energy facilitate its tendency to accept an electron rather than donate it.
The bond between potassium and chlorine that results in KCl is termed an electrovalent bond.
Reaction equation:
K + Cl → KCl
Explanation:
Filtration serves as a method of separation where solid particles that are suspended in a liquid are isolated by passing the mixture through filter paper's pores. This process ensures that the solid particles accumulate on the filter paper and the liquid flows out through the filter paper's pores.
The ordered sequence of the steps provided is:
- Measure and fold the filter paper.
- Insert the filter paper into the funnel, then position the funnel above the Erlenmeyer flask.
- Let the solid/liquid mixture pass through the filter.
- Rinse the filter paper that holds the mixture with water.
- Measure the weight of the dry filter paper along with the copper.
Response:
9.9 ml of 0.200M NH₄OH(aq)
Reasoning:
3NH₄OH(Iaq) + FeCl₃(aq) => NH₄Cl(aq) + Fe(OH)₃(s)
What volume in ml of 0.200M NH₄OH(aq) will fully react with 12ml of 0.550M FeCl₃(aq)?
1 x Molarity of NH₄OH x Volume of NH₄OH Solution(L) = 2 x Molarity of FeCl₃ x Volume of FeCl₃ Solution
1(0.200M)(Volume of NH₄OH Soln) = 3(0.550M)(0.012L)
=> Volume of NH₄OH Soln = 3(0.550M)(0.012L)/1(0.200M) = 0.0099 Liters = 9.9 milliliters
