The ionic formula for Mg and Br is MgBr2, with magnesium having a 2+ charge and bromine having a 1- charge. MgBr2 indicates that two bromide ions combine with one magnesium ion in the compound.
One formula unit of MgBr2 has three ions; one Mg2+ ion and two Br- ions. One mole of MgBr2 formula units has one mole of Mg2+ ions and two moles of Br- ions, for a total of three moles of ions.
There are 0.021 moles of MgBr2 in 3.50 g. Since MgBr2 contains 2 bromide ions per formula unit, there are 0.042 moles of Br- ions. So, there are 0.042 * 6.022 x 10^23 = 2.53 x 10^22 Br- ions in 3.50 g of MgBr2.
There are a total of 2 moles of anions in 2.50 g of MgBr2. Each formula unit of MgBr2 contains 2 moles of anions (Br-). The molar mass of MgBr2 is 184.113 g/mol, so 2.50 g is equivalent to 0.0136 moles, and therefore 0.0136 moles * 2 moles = 0.0272 moles of anions.
The chemical formula for magnesium bromide (MgBr) can be represented in Lewis dot diagram form as follows: Magnesium (Mg) with 2 dots (representing 2 valence electrons) and Bromine (Br) with 7 dots (representing 7 valence electrons) sharing one pair of electrons.
The chemical formula for magnesium bromide is MgBr2. It consists of one magnesium atom bonded to two bromine atoms.
Formula: MgBr2
MgBr2
MgBr2.
The chemical formula for magnesium bromide is MgBr2. This compound is composed of one magnesium ion (Mg2+) and two bromide ions (Br-).
The chemical formula for the compound formed between magnesium and bromine is MgBr2. It is named magnesium bromide.
magnesium bromide
The ionic formula for Mg and Br is MgBr2, with magnesium having a 2+ charge and bromine having a 1- charge. MgBr2 indicates that two bromide ions combine with one magnesium ion in the compound.
MgBr MgBr2
*MgBr2
The answer is 5,978 moles.
The empirical formula for a compound containing 13% magnesium and 87% bromine is MgBr2. This is because the ratio of magnesium to bromine atoms in the compound is 1:2, which corresponds to the formula MgBr2.