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Chemistry, 09.12.2019 18:31 greystokey

Into a 0.25 m solution of nacl(aq), excess al(no3)3(aq) was added to form alcl3(s). 3nacl(aq) + al(no3)3(aq) ? alcl3(s) + 3nano3(aq) if you knew the volume of the solution containing nacl(aq), determine how you would predict the mass of alcl3(s) formed by completing the following solution map. drag the appropriate labels to their respective targets.

2)dimensional analysis simply refers to the inclusion of units in an equation. after setting up the solution map, dimensional analysis can be used to set up the conversion factors that lead to the desired values within this map. the correct setup of an equation can be verified by checking if the result will only have the desired units after unit cancellation. 3nacl(aq) + al(no3)3(aq) ? alcl3(s) + 3nano3(aq) for the described reaction, you have a 0.280 l solution of 0.700 m nacl(aq). using the solution map from part a as a guide, complete the following dimensional analysis that allows you to calculate the moles of alcl3(s) that can be formed by placing the values of each conversion factor according to whether they should appear in the numerator or denominator. drag the appropriate values to their respective targets.

3) when you are presented with additional details, it can sometimes be a little more difficult to determine which information is necessary and how molarity should be applied. using both a solution map and dimensional analysis can clarify which information and conversion factors are necessary to determine the desired value. the following dimensional analysis setup could be used to determine the theoretical mass of albr3(s) (molecular mass = 266.69 g/mol ) produced based on reacting 62.2 g of a 0.016 mol/l solution of br2(l) (density = 1016 g/l ) with excess al(s) as described in the following equation: 3br2(l) + 2al(s) ? 2albr3(s) complete the dimensional analysis for calculating the mass of the product by placing the values of each conversion factor according to whether they should appear in the numerator or denominator when calculating the mass of albr3(s) produced from a sample of br2(l). drag the appropriate values to their respective targets.

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Into a 0.25 m solution of nacl(aq), excess al(no3)3(aq) was added to form alcl3(s). 3nacl(aq) + al(n...
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