Answer:
1.97 moles of Na contains greater number of atoms than 6.8 × 10²² atoms of Li.
Explanation:
To solve this problem we will first calculate the number of atoms contained by 1.97 moles of Na and then will compare it with 6.8 × 10²² atoms of Li.
As we know that 1 mole of any substance contains exactly 6.022 × 10²³ particles which is also called as Avogadro's Number. So in order to calculate the number of atoms contained by 1.97 moles of Na, we will use following relation,
Moles = Number of Atoms ÷ 6.022 × 10²³ atoms.mol⁻¹
Solving for Number of Atoms,
Number of Atoms = Moles × 6.022 × 10²³ Atoms.mol⁻¹
Putting values,
Number of Atoms = 1.07 mol × 6.022 × 10²³ Atoms.mol⁻¹
Number of Atoms = 1.18 × 10²⁴ Atoms of Na
Hence,
1.97 moles of Na contains greater number of atoms than 6.8 × 10²² atoms of Li.
Answer:
O(Oxygen)
Explanation:
Oxygen or O among the given options has the highest electronegativity of 3.5 as compared to 2.4 of sulfur. Thus, there is a lesser attraction in the bonds of Sulfur. <u>The other two elements Al(Aluminium) and Cu are less electronegative than Sulfur and Oxygen as they are metals and possess electropositivity while the latter displays more electronegativity due to being non-metals</u>. Hence, Oxygen is the answer.
Answer:
0.97014 moles KMnO4
Explanation:
1 g KMnO4 = 0.00633 mol
153.26 g x 0.00633 mol/ 1 g KMnO4 = 0.97014 moles KMnO4
