Iodine has the largest heat of vaporization: 41,57 kJ/mol.
a molecule of bromine
Yes, because BrBr2 is ionically bonded, while the Br2O is covalently (molecularly) bonded. Ionic bonds are stronger than covalent (molecular) bonds. This means that BeBr2 has stronger attraction between molecules and it will take more energy (heat) to "unstick them" into the gas phase. Because there is more heat needed to get BeBr2 boiling, it has a higher boiling point.
Hydrogen, Nitrogen, Oxygen, Chlorine and Fluorine are diatomic gases. Bromine is a diatomic liquid. Iodine is a diatomic solid. (at room temperature)
Bromine is nonpolar and needs a nonpolar solvent to do the bromination reaction. CCl4 is a heavy, nonpolar solvent that dissolves the heavy Br2 molecule. With lighter nonpolar solvents such as hexane, the high density Bromine settles out.
The 7 elements that readily form diatomic molecules are hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine, and Iodine. Astatine might theoretically form diatomic molecules, but it is so rare and radioactive that it is hard to study.
Chloroform has a low heat of vaporization due to its relatively weak van der Waals forces between molecules. These weak intermolecular forces require less energy to break apart the molecules and transition from liquid to gas phase. Additionally, chloroform's molecular structure contributes to its low heat of vaporization compared to other compounds.
The balanced equation is: Br2 + 2Kl -> 2KBr + I2. This means there is a 2 in front of the underlined substance, KI.
2 atoms of Bromine[Br] are in Molecular Bromine(Br2)
The pairing of Br2 with polar covalent bonding is incorrect. Br2 exhibits nonpolar covalent bonding due to the similar electronegativities of the two bromine atoms causing a symmetrical distribution of electrons, resulting in a nonpolar molecule.
The reaction 2KBr → 2K + Br2 is endothermic because heat must be supplied to break the bonds in potassium bromide (KBr) in order to form potassium (K) and bromine gas (Br2).
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.
No, the number of atoms in 1 mol of Br2 is equal to Avogadro's number multiplied by 2, because there are 2 atoms of bromine in each molecule of Br2. Avogadro's number represents the number of entities (atoms, molecules, etc.) in 1 mol of a substance.
Br2 + 3NaHSO3 = 2NaBr + NaHSO4 + H2O + 2SO2
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
There are two bromine atoms in Br2
The balanced equation for the reaction between zinc (Zn) and bromine (Br2) is: Zn + Br2 -> ZnBr2.
2 Na + Br2 --> 2 NaBr