Thus problem is providing us with the mass of iron (III) oxide as 12.4 g so the moles are required and found to be 0.0776 mol after the calculations:
<h3>Mole-mass relationships:</h3>
In chemistry, we use mole-mass relationships in order to calculate grams from moles and vice versa. In this case, since we are given the mass of iron (III) oxide as 12.4 g one can calculate the moles by firstly quantifying its molar mass:
Then, we prepare a conversion factor in order to cancel out the grams and thus, get moles:
Learn more about mole-mass relationships: brainly.com/question/18311376
<span>To work out the volume of something from its density, use the compound measures triangle: mass over density and volume. To find volume that the beaker holds, divide the mass by the density. V = (388.15 - 39.09)/1. V = 349.06g/cm3. To find the weight of the beaker and the contents, first work out the weight (mass) of the mercury, with this formula: mass = d x v. M = 13.5 x 349.06. M = 4712.31. Then add on the weight of the beaker (39.09g). The total weight is 4751.40g.</span>
Answer:
both sugar and phosphate molecules
Answer: Many fronts cause weather events such as rain, thunderstorms, gusty winds and A weather front is a transition zone between two different air masses at the Lifted warm air ahead of the front produces cumulus or cumulonimbus clouds
Explanation:
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