Answer: N = 2.78 × 10^23 atoms
There are N = 2.78 × 10^23 atoms in 70g of Au2cl6
Completed Question:
Calculate the number of gold atoms in a 70g sample of gold(III) chloride . Be sure your answer has a unit symbol if necessary, and round it to significant digits
Explanation:
Given:
Molar mass of Au2cl6 = 303.33g/mol
Mass of Au2cl6 = 70g
Number of moles of Au2cl6 = 70g/303.33g/mol = 0.231mol
According to the chemical formula of Au2cl6,
1 mole of Au2cl6 contains 2 moles of Au
Number of moles of Au = 2 × 0.231mol = 0.462mole
There are 6.022 × 10^23 atoms in 1 mole of an element.
Number of Atom of gold in 0.462 mole of gold is:
N = 0.462 mol × 6.022 × 10^23 atoms/mol
N = 2.78 × 10^23 atoms
Answer:
8.58 g/cm3
Explanation:
density = mass / volume
be attention that the question gives u the length of
one side of the cube so u should calculate the volume
of the cube using your information of math .
1st we calculate the volume of the cube :
V of cube = a 3= axaxa
= 0.250 x0.250 x0.250
= 0.0156 cm3
2nd step
density = mass/volume
= 0.134/ 0.0156 = 8.58 g/cm3
The atomic number of Fluorine is 9
Valence (outer) electron configuration is : 2s²2p⁵
Therefore, it requires 1 electron in the p-orbital to complete its octet of 8 electrons.
Thus, the atom Fluorine generally will become <u>more </u>stable through the formation of an ionic chemical compound by accepting <u>1 </u> electron from another atom. This process will fill its outer energy level.
Ans: A) more, 1
