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This is because marshmallows are made of sugar and water, while lead is a heavy metal. This means that the sample with the largest mass of 1 mole of marshmallows would be the one with the greatest number of marshmallows. PPT I. Molar Conversions - Copley 1 See answer Advertisement Advertisement yavcak156 yavcak156 Answer: B. fjords, norway cruise capital radiology 157 scoresby road boronia capital radiology 157 scoresby road boronia All information published on this website is provided in good faith and for general use only. john pawlowski obituary; how to prevent albinism during pregnancy; honeyglow pineapple vs regular pineapple; nickelodeon live show tickets; goway travel liquidation Yes, Marshmallow Fluff weighs the same as regular marshmallows. This can be done by using the density of marshmallows. Menu. How do I change the sample rate in pro tools? which sample has the largest mass 1 mole of marshmallowskristen wiig daughter. This is because marshmallows are not very dense. This can be done by using theAvogadro's number. PQ-1. Which sample has the largest mass?A) 1 mole of marshmallowsB) 1 mole of Although this represents just a tiny fraction of 1 mole of water (~18 g), it contains more water molecules than can be clearly imagined. Get started with your FREE initial assessment!https://glasertutoring.com/contact/#MolesToMolecules #MolesToGrams #OpenStaxChemistry Avogadro's number, #6.022 * 10^(23)#. How many hemoglobin molecules is this? Which sample has the largest mass? Determine the number of moles of each component. To do this, we simply need to multiply the mass of one marshmallow by the Avogadro's number. In other words, a single marshmallow has close to 30% more volume than an entire chunk of lead! PDF CHEM 1411 Chapter 2 Homework Answers - austincc.edu Explain why. Likewise, in order to have have one mole of water, you need to have #6.022 * 10^(23)# molecules of water. Since the amount of Ar is less than 1 mole, the mass will be less than the mass of 1 mole of Ar, approximately 40 g. The molar amount in question is approximately one-one thousandth (~103) of a mole, and so the corresponding mass should be roughly one-one thousandth of the molar mass (~0.04 g): In this case, logic dictates (and the factor-label method supports) multiplying the provided amount (mol) by the molar mass (g/mol): [latex]9.2\times {10}^{-4}\cancel{\text{mol}}\text{Ar}\left(\frac{39.95\text{g}}{\cancel{\text{mol}}\text{Ar}}\right)=0.037\text{g Ar}[/latex].