To find the grams of bromine (Br) in 595 g of calcium bromide (CaBr2), first determine the molar mass of CaBr2. The molar mass is approximately 40.08 g/mol for Ca and 79.90 g/mol for Br, giving a total of about 199.89 g/mol for CaBr2. Since there are two bromine atoms in each formula unit, the mass of bromine in CaBr2 is about 159.80 g (2 × 79.90 g). Thus, the mass of Br in 595 g of CaBr2 can be calculated as follows: (159.80 g Br / 199.89 g CaBr2) × 595 g CaBr2 ≈ 477.23 g of Br.
The compound CaBr2 is called calcium bromide. It is an inorganic salt consisting of calcium cations (Ca2+) and bromide anions (Br-). It is commonly used in various industrial applications.
2,60x102 grams of bromine (Br) is equal to 1,627 moles Br2.
CaBr2 would produce a stronger ionic lattice structure compared to BaBr2. This is due to the higher charge density of calcium ions (Ca²⁺) compared to barium ions (Ba²⁺), as calcium has a smaller ionic radius. The stronger electrostatic forces between the Ca²⁺ ions and Br⁻ ions lead to a more stable and stronger lattice structure in CaBr2.
To find the mass of 0.0751 mol of Br₂ (bromine), you first need the molar mass of Br₂, which is approximately 159.808 g/mol (since the atomic mass of bromine is about 79.904 g). You can calculate the mass using the formula: mass = moles × molar mass. Thus, the mass of 0.0751 mol Br₂ is approximately 0.0751 mol × 159.808 g/mol = 12.00 grams.
There is one mole in approximately 80 grams of bromine (Br). Therefore, in 1 gram of bromine, there would be approximately 1/80 mole, which is equal to 0.0125 moles.
1st you must find the molar mass of CaBr2. Ca 40.08g * (number of moles in the compound) 1 Br 79.90g * 2 This give you the amount (g) in 1 mole of CaBr2. Multiply by 14 and you get the answer.
All you have to do is flip the numbers and reduce. If you have Ca+2 and Br-, Switch the numbers so you get CaBr2. 1 and 2 do not have any common factors so you don't have to reduce! CaBr2 is the final answer.
In an aqueous solution of CaBr2, the solute particles are Ca2+ cations and Br- anions. When dissolved in water, CaBr2 dissociates into these ions, which are responsible for conducting electricity and other properties of the solution.
The formula for calcium bromide is CaBr2, where Ca represents calcium and Br represents bromine.
The ionic compound calcium bromide is CaBr2.
A solution of CaBr2 is neutral. When CaBr2 dissolves in water, it dissociates into calcium ions (Ca2+) and bromide ions (Br-), which do not significantly affect the pH of the solution.
In this reaction, Br is the reducing agent because it is being oxidized from Br- to Br2, thereby causing the reduction of Ca from Ca to Ca2+.
Ca+2 + Br-1 --> CaBr2 (calcium bromide)
Calcium bromide is the ionic compound represented by CaBr2. It consists of calcium (Ca) cations and bromide (Br) anions, which form a lattice structure in a solid state.
The formula for the ionic compound formed by calcium (Ca2+) and bromine (Br-) would be CaBr2. The calcium ion has a charge of 2+ while the bromine ion has a charge of 1-, so two bromine ions are needed to balance the charge of one calcium ion.
CaBr2
The compound CaBr2 is called calcium bromide. It is an inorganic salt consisting of calcium cations (Ca2+) and bromide anions (Br-). It is commonly used in various industrial applications.