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What is the empirical formula of the ionic compound formed by Ca2 and Br?
The empirical formula is CaBr2. This is because calcium has a 2+ charge (Ca2+) and bromine has a 1- charge (Br-), so the compound would need two bromine ions to balance the charge of one calcium ion.
What is the empirical formula for Al2Br6?
it is dimeric form of Aluminium bromide AlBr3
What is the fromula for bromide?
Formula: Br-
What is the empirical formula for a compound that is 7.9 percent Li and 92.1 percent Br?
The atomic mass of lithium is 6.94, while that for bromine is 79.90. 7.9/6.94 = 1.095 and 92.1/79.90 = 1.15. the closest integer ratio between these two number si 1:1, so the empirical formula is LiBr.
What is the empirical formula of 22 percent Carbon 4.6 percent Hydrogen 73.4 percent Bromine?
To determine the empirical formula, we first need to convert the percentages to moles. Assuming we have 100g of the compound, we have 22g C, 4.6g H, and 73.4g Br. Next, we calculate the moles of each element: 22g C / 12.01 g/mol = 1.83 mol C, 4.6g H / 1.01 g/mol = 4.55 mol H, and 73.4g Br / 79.9 g/mol = 0.92 mol Br. Finally, we divide each mole value by the smallest mole value to get the simplest ratio, which gives us the empirical formula: C1H2Br1, or CH2Br.
What is the empirical formula of the ionic compound formed by Ca2 and Br?
The empirical formula is CaBr2. This is because calcium has a 2+ charge (Ca2+) and bromine has a 1- charge (Br-), so the compound would need two bromine ions to balance the charge of one calcium ion.
What is the empirical formula for a compound if a 20.0 g sample contains 4.0 g Ca and 16.0 g Br?
CaBr2
What is the molecular formula of a compound if its empirical formula is CFBrO and its molar mass is 381.01 gmol?
To determine the molecular formula from the empirical formula CFBrO, we first calculate the molar mass of the empirical formula: C (12.01 g/mol) + F (19.00 g/mol) + Br (79.90 g/mol) + O (16.00 g/mol) = 126.91 g/mol. Next, we divide the given molar mass (381.01 g/mol) by the empirical formula mass (126.91 g/mol) to find the ratio: 381.01 g/mol ÷ 126.91 g/mol ≈ 3. The molecular formula is thus (CFBrO)₃, or C₃F₃Br₃O₃.
What is the empirical formula for calcium bromide dihydrate?
The empirical formula for calcium bromide dihydrate is CaBr2·2H2O. This means there are two bromide (Br) ions for every calcium (Ca) ion, and two water (H2O) molecules per unit.
What is the empirical formula and molecular formula of a compound with a molar mass of 245.8g The composition is 19.53 C 2.44 H 13.02 O and 65.01 Br?
The molar mass of the empirical formula is calculated by summing up the molar masses of the elements in the given composition (which gives a molar mass of 281.6 g/mol). To find the empirical formula, divide the molar mass of the compound (245.8 g/mol) by the molar mass of the empirical formula (281.6 g/mol), which gives approximately 0.873. This means the empirical formula is BrC₆H₈O₃.
What is the empirical formula for Al2Br6?
it is dimeric form of Aluminium bromide AlBr3
What is the fromula for bromide?
Formula: Br-
When 8.95g of hydrated magnesium bromide is heated 2.03g of water is driven off What is the empirical Formula?
To find the empirical formula of hydrated magnesium bromide, we first determine the moles of magnesium (Mg), bromine (Br), and water (H₂O) after heating. The mass of anhydrous magnesium bromide is 8.95g - 2.03g = 6.92g. The moles of water lost (2.03g) is calculated as 2.03g / 18.02g/mol = 0.112 moles. Assuming magnesium bromide has the formula MgBr₂, we find the moles of Mg and Br in the remaining mass, leading to an empirical formula of MgBr₂·xH₂O, where x is determined by the ratio of moles of water to moles of MgBr₂.
What is the empirical formula for a compound that is 7.9 percent Li and 92.1 percent Br?
The atomic mass of lithium is 6.94, while that for bromine is 79.90. 7.9/6.94 = 1.095 and 92.1/79.90 = 1.15. the closest integer ratio between these two number si 1:1, so the empirical formula is LiBr.
What is the empirical formula of 22 percent Carbon 4.6 percent Hydrogen 73.4 percent Bromine?
To determine the empirical formula, we first need to convert the percentages to moles. Assuming we have 100g of the compound, we have 22g C, 4.6g H, and 73.4g Br. Next, we calculate the moles of each element: 22g C / 12.01 g/mol = 1.83 mol C, 4.6g H / 1.01 g/mol = 4.55 mol H, and 73.4g Br / 79.9 g/mol = 0.92 mol Br. Finally, we divide each mole value by the smallest mole value to get the simplest ratio, which gives us the empirical formula: C1H2Br1, or CH2Br.
What is the chemical formula of Bromides?
Formula: Br-
What is the complete structural formula for 1-2-Dibromoethane?
Br Br | |H - C - C - H| |H H
