Its nonpolar :)
14.17 mol BaBr2 has 2*14.17 mol Br in it, so 28.34 mol KBr can be produced (also 28.34 mol K is needed)
BaBr2 + 2HCl >> BaCl2 + 2HBr
BaBr2 does not contain any covalent bonds. It is an ionic compound composed of barium cations (Ba2+) and bromide anions (Br-), which are held together by ionic bonds formed through the transfer of electrons.
Yes, BaBr2 is an ionic compound. It consists of barium (Ba), a metal, and bromine (Br), a nonmetal. In BaBr2, barium loses two electrons to become a cation (Ba2+), while bromine gains one electron from each barium atom to form an anion (Br-), resulting in an ionic bond between them.
2 KBr + BaI2 ----> 2 KI + BaBr2
BaBr2
The skeleton equation for the reaction between barium and bromine would be: Ba + Br2 -> BaBr2.
14.17 mol BaBr2 has 2*14.17 mol Br in it, so 28.34 mol KBr can be produced (also 28.34 mol K is needed)
BaBr2 + 2HCl >> BaCl2 + 2HBr
BaBr2 does not contain any covalent bonds. It is an ionic compound composed of barium cations (Ba2+) and bromide anions (Br-), which are held together by ionic bonds formed through the transfer of electrons.
Yes, BaBr2 is an ionic compound. It consists of barium (Ba), a metal, and bromine (Br), a nonmetal. In BaBr2, barium loses two electrons to become a cation (Ba2+), while bromine gains one electron from each barium atom to form an anion (Br-), resulting in an ionic bond between them.
14.17 mol BaBr2 has 2*14.17 mol Br in it, so 28.34 mol KBr can be produced (also 28.34 mol K is needed)
2 KBr + BaI2 ----> 2 KI + BaBr2
No, BaBr2 is a crystalline solid compound that forms a specific lattice structure, making it difficult to shape or mold into different forms. Its rigid structure gives it a fixed characteristic shape at the atomic level.
Barium bromide is an inorganic compound composed of barium and bromine, with the chemical formula BaBr2. It is a white crystalline solid and is commonly used in a variety of applications, including as a precursor in organic synthesis and in the production of specialty glass.
None, unless there is metallic potassium in the reaction mixture. Assuming excess potassium metal is present then 14 moles of KBr can be produced. 7BaBr2 + excess potassium -----> 14KBr + 7 Ba
Unbalanced: KBr + BaI2 --> KI + BaBr2Balanced: 2KBr + BaI2 --> 2KI + BaBr2