The average atomic mass of an element can be determined by multiplying the individual masses of the isotopes with their respective relative abundances, and adding them.
Average atomic mass of Br = 158 amu(0.2569) + 160 amu(0.4999) + 162 amu(0.2431)
Average atomic mass = 159.96 amu
As described in the problem, the relative abundance for Br-79 is 25.69%. This is because 2 atoms of Br is equal to 79*2 = 158 amu. Similarly, the relative abundance of Br-81 is 81*2 = 162, which is 24.31%.
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: Metalloids
Explanation:
The dark “stair step” line on the right side of the periodic table is where the metalloids are. Metalloids “touch” the SSL. (Stair step line)
The metals are to the left of the SSL and the non metals are to the right of the SSL.
Answer:
for the process is -375 kJ
Explanation:
- Given reaction is a combination of the two given elementary steps.
- Summation of change in standard enthalpy ()of the two elementary reactions give of the reaction .
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