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Sodium bromide (NaBr) has a weight of 102.89 grams per mole. Determining the mass of NaBr in a mixture can be calculated if the constituent weights of all the others are known. The total masses of the knows can be subtracted by the overall weight to isolate the mass of NaBr.

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Which is the percent composition of bromine in the compound NaBr?

The percent composition of bromine in NaBr is approximately 77.7%. This is calculated by dividing the molar mass of bromine by the molar mass of NaBr and then multiplying by 100.


What is the percent composition of bromine in the compound NaBr?

Bromine has an atomic mass of 79.904 g/mol, while the molar mass of NaBr is 102.89 g/mol. Therefore, the percent composition of bromine in NaBr is (79.904 g/mol / 102.89 g/mol) x 100 = 77.68%.


What is the mass of NaBr that will be produced from 42.7 g of AgbR?

Na2S2O3 + AgBr → NaBr + Na3[Ag(S2O3)2] First check that the given equation is balanced ... it isn't ... so the first thing to do is balance the equation: balancing Na: 2Na2S2O3 + AgBr → NaBr + Na3[Ag(S2O3)2] and everything is now balanced so we've got the balanced equation molar mass AgBr = 107.87 + 79.90 = 187.77 g/mol mol AgBr available = 42.7 g AgBr x [1 mol / 187.77 g] = 0.2274 mol AgBr from the balanced equation the mole ratio AgBr : Na2S2O3 = 1 : 2 so mol Na2S2O3 required = 0.2274 mol AgBr x [ 2 mol Na2S2O3 / mol AgBr] = 0.455 mol Na2S2O3 (to 3 sig figs)


A mixture of NaCl and NaBr has a mass of 2.03 g and is found to contain 0.76 grams of Na What is the mass of NaBr in the mixture?

To find the mass of NaBr in the mixture, we first calculate the mass of Na in NaCl and NaBr combined, which is 0.76 g. Since NaCl contains one Na atom and NaBr contains one Na atom, this entire mass comes from Na in NaCl and NaBr. Therefore, the mass of NaCl is 0.76 g and the remaining mass of the mixture (2.03 g - 0.76 g = 1.27 g) is due to NaBr.


How many grams of NaBr are contained in 75.0 mL of a 1.5 M NaBr solution?

Molarity = moles of solute/Liters of solution ( 75.0 ml = 0.075 Liters ) Get moles NaBr 1.5 M NaBr = moles NaBr/0.075 Liters = 0.1125 moles NaBr (102.89 grams/1 mole NaBr) = 11.575 grams NaBr ( call it 12 grams ) ----------------------------------------------------


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How many grams of sodium does a 5.35 g sample of sodium bromide contain?

All you really need here is a ratio of molar masses, ie molar mass of Na/molar mass of NaBr So look at your periodic table and is says: Na = 23g/mol Br = 80g/mol therefore NaBr = 103g/mol so from this, you actually know a percentage of Sodium to Bromine, correct me if I'm wrong, someone else please but just take: (23g/mol)/103g/mol = 22.3% of you NaBr is actually Na, so take 5.35g * 0.223 and you get: 1.19g of Na in your NaBr Hope that helps. Cheers


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What reactant is limiting if 3000 cm3 of Cl2 at STP react with a solution containing 25 grams of NaBr?

Cl2 + 2NaBr => Br2 + 2NaCl One mole Cl2 reacts with 2 moles NaBr Cl2 = 71 NaBr = 102.9 Molar volume = 22.414 L/mole for ideal gas @STP 3L Cl2 = 3/22.414 = 0.1338 mole 25g NaBr = 25/102.9 = 0.2430 mole 0.1338 moles Cl2 requires 0.2676 moles NaBr for complete reaction The NaBr is the limiting reagent