asked by @valentinac9 •
almost 2 years ago •
Chemistry
• 5 pts

various “coin ion” combinations, assuming the dime to be the cation. With actual coins, determine the number of anions that can touch the dime with each combination. I really don't know where to start

Hey there. Interesting question. In this question we need to determine how many coins (pennies, quarter, and nickels) you can put around the circumference of a dime. So we have diameters given and so recall the radius is the diameter divided by 2.

The radius of a dime would be 17/2 = 8.5 nm. We are told this is the cation so it's r+ The radius of a penny would be 19/2 = 9.5 nm. Make this r- penny The radius of a nickel would be 21/2 = 10.5 nm. Make this r- nickel. The radius of a quarter would be 24/2 = 12.0 nm. Make this r- quarter.

So now divide r+ by the r- of each coin to determine the different combinations.

dime/penny = r+/r- = 8.5/9.5 = 0.895 dime/nickel = r+/r- = 8.5/10.5 = 0.810 dime/quarter = r+/r- = 8.5/12.0 = 0.708

This gives us

No. of penny anions that can touch dime cation = 6

No. of nickel anions that can touch dime cation = 5

No. of quarter anions that can touch dime cation = 5

answered by @jules •
almost 2 years ago

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