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
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Answer:
The arrangement of the periodic table leads us to visualize certain trends among the atoms.
Elements are arranged from left to right and top to bottom in order of increasing atomic number.
The vertical columns (groups) of the periodic table are arranged so that all its elements have the same number of valence electrons (outer ring). All elements within a certain group share similar properties.
The rows are called periods. All elements in a row have the same number of electron shells.
First, find moles of oxygen gas: (3.01 x10^23 molec.)/(6.02 x10^23) =0.5mol O2
Second, multiply moles by the standard molar volume of a gas at STP:(0.5mol)(22.4L) = 11.2L O2
