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Q: How many moles are contained in 44.0 g Br2?
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How many moles of atoms are in 44.0 Br2?

44.0 grams Br2 ? 44.0 grams Br2 (1 mole Br2/159.8 grams)(6.022 X 10^23/1 mole Br2)(1 mole Br2 atoms/6.022 X 10^23) = 0.275 moles of Br2 atoms


How many particles are in 9.00 moles of Br2?

9 moles of bromine contain 54,2.10e23 molecules.


How many grams are in 2.9 moles of Br2?

2,9 moles of bromine is equivalent to 463,4432 g.


How many moles of Br2 are there in 160g?

1,012 mole of bromine for the diatomic molecule.


How many moles of bromine will be formed upon the complete reaction of 0.248 moles bromine trifluoride?

The complete decomposition reaction is as follows:2 BrF3 → Br2 + 3 F2 , so 2 moles BrF3 will give 1 mole Br2 , hence 0.248 mole gives 0.124 mole Br2


How do you calculate the number of moles of bromine present in 16.0ml of Br2 whose density is 3.12gml?

First convert the volume of the Br2 into grams by using:D=M/VSo we are given that volume=16.0 ml and density=3.12g/ml.M=D*VM=(3.12g/ml)*(16.0ml)=49.92 gThen we use #moles of a substance=#grams present/Formula weight(# of grams of Br2 in 1 mol of Br2)The Formula weight(molar mass) of Br2=2*(79.9 g/mol)=159.80 g/mol Br2#moles of Br2=49.92g/159.80g/mol Br2=.312 moles of Br2 present.


How many atoms are in 1.54 moles of Bromine?

1.54 (mol Br2) * 6.022*10+23 (molecule/mol Br2) * 2 (atoms Br/molecule Br2) =1.85*1024 atoms in 1.54 mole Br2


How many moles of bromine atoms are in 2.60x102 grams of bromine?

2,60x102 grams of bromine (Br) is equal to 1,627 moles Br2.


What is the mass of 0.030 mole of Br2?

The mass of 0.030 moles of Br2 is 4.79424 grams, properly rounded to 4.8 grams.


How many atoms are in 1.54 moles of Bromine Br?

If it is 1.54 moles of Br atoms then the answer is 9.274 X 1023 atoms.If it is 1.54 moles of Br2 molecules then the answer is 1.855 X 1024 atoms.


How many moles of lithium chloride will be formed by the reaction of chlorine with 046 moles of lithium bromide in the following reaction 2 LiBr plus Cl2 equals 2 LiCI plus Br2?

2LiBr(aq) + Cl2(g) = 2LiCl(aq) + Br2(l) will result in .167 moles of lithium chloride.


How many moles of potassium bromide can be produced from the reaction of 2.92 moles of potassium with 1.78 moles of bromine gas?

First write a balanced chemical equation: 2K + Br2 ---> 2KBR Find the limiting reactant by using the moles of each element and determining which one gives you the smallest number of moles of potassium bromide. 2.92 mol K (2 mol KBr/2 mol K)= 2.92 mol KBr 1.78 mol Br2 (2 mol KBR/1 mol Br2)=3.56 mol KBr potassium is your limiting reactant so the max. number of moles of KBr that can be produced is 2.92 mol of KBr