SrCl2 Strontium (Sr) = 87.62 grams/mol 2 Chlorine (Cl) = 70.9 grams/mol ----------------------------------------------------add Strontium chloride = 158.52 grams/mol ============================
The answer is 0,3422 grams.
92.4 grams
This depends on the chemical compound.
1 Mol
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.
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
1 mole Br2 = 159.808g Br2 = 6.022 x 1023 molecules Br2 4.89 x 1020 molecules Br2 x 1mol Br2/6.022 x 1023 molecules Br2 x 159.808g Br2/mol Br2 = 0.130g Br2
SrCl2 Strontium (Sr) = 87.62 grams/mol 2 Chlorine (Cl) = 70.9 grams/mol ----------------------------------------------------add Strontium chloride = 158.52 grams/mol ============================
The answer is 0,3422 grams.
92.4 grams
769.0 grams
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
357
As the molar mass of NaCl is 58.5 g/mol, 1.6 moles weight is 93.6 grams.
No the number of ATOMS in 1 Bomine MOLECULE Br2 is twice Avagadro's number.
This depends on the chemical compound.