Object a: has a mass of 2g a volume of .5lobject b: has a mass of 3g and a volume of 5lwater has a density of 1g/l, which object will sink in water? which one will float? i will you20 points if correct

At 40 âc the solution has at 40 â c the solution has blank g of k n o 3 per 100 g of water and it can contain up to blank g of k n o 3 per 100 g of water. at 0 â c the solubility is ~ blank g k n o 3 per 100 g of water, so ~ blank g k n o 3 per 100 g of water will precipitate out of solution.g of kno3 per 100 g of water and it can contain up to at 40 â c the solution has blank g of k n o 3 per 100 g of water and it can contain up to blank g of k n o 3 per 100 g of water. at 0 â c the solubility is ~ blank g k n o 3 per 100 g of water, so ~ blank g k n o 3 per 100 g of water will precipitate out of solution.g of kno3 per 100 g of water. at 0 âc the solubility is ~ at 40 â c the solution has blank g of k n o 3 per 100 g of water and it can contain up to blank g of k n o 3 per 100 g of water. at 0 â c the solubility is ~ blank g k n o 3 per 100 g of water, so ~ blank g k n o 3 per 100 g of water will precipitate out of solution.kno3 per 100 g of water, so ~ at 40 â c the solution has blank g of k n o 3 per 100 g of water and it can contain up to blank g of k n o 3 per 100 g of water. at 0 â c the solubility is ~ blank g k n o 3 per 100 g of water, so ~ blank g k n o 3 per 100 g of water will precipitate out of solution.gkno3 per 100 g of water will precipitate out of solution. a kno3 solution containing 55 g of kno3 per 100.0 g of water is cooled from 40 ∘c to 0 ∘c. what will happen during cooling?

Which diagram represents the phase tha occurs after a solid melts?

Which of the following represents a balanced nuclear equation showing bismuth-212 undergoing alpha decay followed by beta decay? (2 points) superscript 212 over subscript 83 bi yields superscript 4 over subscript 2 he + superscript 208 over subscript 81 tl yields x−superscript 208 over subscript 82 pb + superscript 0 over subscript -1 beta superscript 212 over subscript 83 bi + superscript 0 over subscript -1 beta yields superscript 4 over subscript 2 he + superscript 208 over subscript 81 tl yields x−superscript 208 over subscript 82 pb superscript 216 over subscript 85 as yields superscript 4 over subscript 2 he + superscript 216 over subscript 85 as yields superscript 216 over subscript 85 pb + superscript 0 over subscript -1 beta superscript 212 over subscript 83 bi yields superscript 0 over subscript -1 beta + superscript 208 over subscript 81 tl yields superscript 208 over subscript 82 pb + superscript 4 over subscript 2 he