Answer:
56.17% is percent composition of Sb in the molecule
Explanation:
Percent composition is the percent in mass of each element present in a particular molecule.
In SbF₅, there is 1 mole of Antimony -Molar mass: 121.76g- per 5 moles of fluorine -Molar mass: 19g/mol-. In a basis of 1 mole, the mass of Sb and F is:
<em>Mass Sb:</em>
1mol * (121.76g/mol) = 121.76g
<em>Mass F:</em>
1 mol SbF₅ = 5 moles F * (19g / mol) = 95g
<em>Total mass:</em>
121.76g + 95g = 216.76g
And percent composition of Sb:
121.76g / 216.76g * 100 =
<h3>56.17% is percent composition of Sb in the molecule</h3>
Answer:
Explanation:
molar mass of BaCl2 = 208.23
mol of BaCl2 = 8/208.22
mol of BaCl2 = 0.03841905
Molarity = 0.03841905/0.450
Molarity = 0.085 M
Answer: After three half-lives 1/8 (12.5%) of the original sample remains
The oxidation half-reaction occurs at one electrode (the anode), and the reduction half-reaction occurs at the other (the cathode). When the circuit is closed, electrons flow from the anodeto the cathode.
