Three
ionic bonding
Al(OH)3 + 3NaNO3 >> Al(NO3)3 + 3NaOH This is a double displacement reaction between metals ( Aluminum is a transition metal, but has only a 3+ oxidation state. ) and polyatomic ions of negative charge. Both are 1- in charge. Sodium is 1+ in charge. This is why the atoms are arranged as they are with the polyatomic ions enclosed in parentheses where needed. Remember, botgh sides of the equation musr contain the same number of atoms. Count carefully and you will see this is the case.
To find the volume of calcium hydroxide solution needed to react with the phosphoric acid solution, you need to determine the mole ratio between calcium hydroxide and phosphoric acid. The balanced chemical equation for the reaction will guide you in calculating the amount needed. Once you have the mole ratio, you can use the concentrations and volumes of the solutions to determine the volume of calcium hydroxide needed.
Sodium chloride is neutral.
equal number of each ion in the compound. This is because potassium (K) typically forms a +1 ion and bromine (Br) typically forms a -1 ion, so one of each is needed to balance the charges and create a neutral compound.
The aluminum ion has a 3+ charge, while hydroxide ions have a 1- charge. To form a neutral compound, the charges must balance. Three hydroxide ions (3 x 1-) are needed to balance the 3+ charge of the aluminum ion, resulting in Al(OH)3, which is neutral.
To prepare alum from scrap aluminum, begin by cutting the aluminum into small pieces and dissolving it in a sodium hydroxide solution. Filter out any impurities, then add sulfuric acid to the solution to precipitate aluminum hydroxide. Lastly, treat the aluminum hydroxide with sulfuric acid and allow the solution to evaporate to obtain alum crystals.
The ionic compound for iodine and aluminum is aluminum iodide, with the chemical formula AlI3. Aluminum has a 3+ charge and iodine has a 1- charge, so three iodine atoms are needed to balance the charge of one aluminum atom.
To find the amount of aluminum needed to produce aluminum sulfate, you need to consider the molar mass of aluminum sulfate and the ratio of aluminum in the compound. First, calculate the molar mass of aluminum sulfate (Al2(SO4)3). Then, find the ratio of aluminum in the compound (2 moles of Al in 1 mole of Al2(SO4)3). Finally, use this information to calculate the grams of aluminum needed to produce 25.0 grams of aluminum sulfate.
ionic bonding
Fe 3+andOH -form the ionic compoundFe(OH)3-------------------------ferric hydroxide, or the modern name, iron (III) hydroxide.
The ionic formula for Aluminum fluoride is AlF3. In this compound, aluminum has a 3+ charge and fluoride has a 1- charge, so to balance the charges, three fluoride ions are needed for each aluminum ion.
The correct formula for aluminum chloride is AlCl3. This is because aluminum typically forms a 3+ ion (Al3+) and chloride forms a 1- ion (Cl-), so three chloride ions are needed to balance the charge of one aluminum ion in the compound.
One of each K+ + I- -----> KI.
The sulfate ion (SO4^2-) forms a neutral compound when combined with a group 1a monatomic ion in a 1:2 ratio. This means that for every one monatomic ion from group 1a, two sulfate ions are needed to form a neutral compound.
The number of hydroxide ions needed in a chemical formula is determined by the charge of the cation in the compound. The hydroxide ion has a charge of -1, so the number of hydroxide ions required is equal to the absolute value of the cation's charge.
Here is a procedure for isolating a neutral compound from a mixture containing basic impurities: Begin by adding an acid to the mixture containing the neutral compound and basic impurities. This will convert the basic impurities into their protonated forms which can be extracted with the aqueous layer. Perform liquid-liquid extraction using a solvent that is immiscible with water. The neutral compound will partition into the organic solvent while the protonated basic impurities will remain in the aqueous layer. Separate the organic layer containing the neutral compound and evaporate the solvent to obtain the pure compound. Further purify the compound using techniques such as recrystallization or chromatography if needed.