Chemistry Help Forum Moles as Intermediate Questions
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 Aug 17th 2018, 10:29 AM #1 ayoolu11 Junior Member   Join Date: Aug 2018 Location: Saskatchewan Posts: 1 Moles as Intermediate Questions What mass of CuBr is found in 4 L of 0.432 mol/L solution ? I n the boxes below place the following information: the moles of CuBr the molar mass of CuBr the mass of CuBr What mass of KNO3 is found in 0.44 L of 1.55 mol/L solution ? I n the boxes below place the following information: the moles of KNO3 the molar mass of KNO3 the mass of KNO3 How many atoms of Hg are found in 662 g ? In the boxes below place the following information: the molar mass of Hg the moles of Hg the number of atoms of Hg How many atoms of Ag are found in 24 g ? I n the boxes below place the following information: the molar mass of Ag the moles of Ag the number of atoms of Ag What volume of CO2 gas is found in 466 g ? In the boxes below place the following information: the molar mass of CO2 the moles of CO2 the volume of CO2 gas What volume of N2 gas is found in 0.099 g ? In the boxes below place the following information: the molar mass of N2 the moles of N2 the volume of N2 gas URGET HELP PLEASE!
 Sep 7th 2018, 04:24 AM #2 Woody Member   Join Date: Sep 2018 Location: England Posts: 50 Possibly too late for an URGENT reply. First I would like to make the point that this forum is not here to do your homework for you. However if you don't understand the homework question, we are willing to attempt to explain it. Firstly are you happy with what a Mole is? 1 mole is a very big (mind bogglingly huge) number (sometimes called Avagadro's Number) about 600,000,000,000,000,000,000,000 (there are 23 zeros) So one litre of a 0.432 mol/L solution contains: 0.432*600,000,000,000,000,000,000,000 molecules of the dissolved substance (CuBr for example). So 4 Litres contains 4*0.432*600,000,000,000,000,000,000,000 molecules of the dissolved substance Now happily chemists who have gone before us have spent a great deal of time and effort to determine how much 600,000,000,000,000,000,000,000 atoms of each element weighs and have made tables of the results. For Copper 600,000,000,000,000,000,000,000 atoms weigh 63.546 grams For Bromine 600,000,000,000,000,000,000,000 atoms weigh 79.904 grams So for CuBr 600,000,000,000,000,000,000,000 molecules weigh 63.546+79.904 grams so the answer is 4*0.432*(63.546+79.904) grams If you understand this one, the other questions are just a slightly different combination of the same ideas. Ok Avogadro's number is 6.023.... × 10^23, I rounded it off a bit for simplicity. But note that the actual number does not appear in the final answer, so is it not vital to know what the number actually is! You just need to know that these weights of different compounds (as read from the aforementioned tables) contain the same number of atoms (or molecules) (within the accuracy to which we can measure).

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