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Balanced equation: LiOH + HBr ---> LiBr + H₂O Here, we aim to convert the mass of LiOH to mass of LiBr. In this formula, the product (LiBr) takes x, and the reactant (LiOH) takes y. Here's how it goes. (? = coefficient in the balanced equation) mass of x = (mole of y) * (? mol x / ? mol y) * (molar mass of x) mass of LiBr = (10 g / 23.95 g/mol) * (1 mol LiBr / 1 mol LiOH) * (86.85 g/mol LiBr) mass of LiBr = 36.3 g (Answer)
To determine the number of grams of lithium nitrate needed to make 250 grams of lithium sulfate, you need to calculate the molar mass of lithium sulfate and lithium nitrate, then use stoichiometry to find the ratio of lithium nitrate to lithium sulfate. Finally, apply this ratio to find the mass of lithium nitrate needed for the reaction. Lead sulfate is not involved in this calculation as it is not part of the reaction between lithium nitrate and lithium sulfate.
Approximately 56.7 grams of potassium bromide will dissolve in 100g of water at 70°C.
Lithium chloride is not transformed in calcium chloride.
To find out the grams of lithium nitrate needed, you need to calculate the molar mass of lithium sulfate (Li2SO4) and lithium nitrate (LiNO3). Then use stoichiometry to determine the amount of lithium nitrate required to produce 250 grams of lithium sulfate. The balanced chemical equation for the reaction would also be needed.
The formula for lithium bromide is LiBr. The compound has a molar mass of 86.845 grams per mole. One of its main uses is as a desiccant.
To determine how many liters of a 4M lithium bromide (LiBr) solution can be made from 100 grams of LiBr, we first need to calculate the number of moles in 100 grams. The molar mass of lithium bromide is approximately 86.84 g/mol, so 100 grams corresponds to about 1.15 moles. A 4M solution contains 4 moles of solute per liter, thus 1.15 moles can produce approximately 0.29 liters (1.15 moles ÷ 4 moles/L). Therefore, 100 grams of lithium bromide can make about 0.29 liters of a 4M solution.
The molecular weight of Calcium Bromide is extremely close to 200. So 1200/200 = 6 moles present.
Balanced equation: LiOH + HBr ---> LiBr + H₂O Here, we aim to convert the mass of LiOH to mass of LiBr. In this formula, the product (LiBr) takes x, and the reactant (LiOH) takes y. Here's how it goes. (? = coefficient in the balanced equation) mass of x = (mole of y) * (? mol x / ? mol y) * (molar mass of x) mass of LiBr = (10 g / 23.95 g/mol) * (1 mol LiBr / 1 mol LiOH) * (86.85 g/mol LiBr) mass of LiBr = 36.3 g (Answer)
At 10°C, the solubility of strontium bromide (SrBr₂) in water is approximately 3.5 grams per 100 grams of water. This means that you can dissolve about 3.5 grams of strontium bromide in 100 grams of water at this temperature.
No, a AA lithium battery has about 0.75 grams of Lithium.
To determine the number of grams of lithium nitrate needed to make 250 grams of lithium sulfate, you need to calculate the molar mass of lithium sulfate and lithium nitrate, then use stoichiometry to find the ratio of lithium nitrate to lithium sulfate. Finally, apply this ratio to find the mass of lithium nitrate needed for the reaction. Lead sulfate is not involved in this calculation as it is not part of the reaction between lithium nitrate and lithium sulfate.
It is not a change at all. 79 grams of lithium is a quantity of a substance.
Approximately 56.7 grams of potassium bromide will dissolve in 100g of water at 70°C.
Lithium chloride is not transformed in calcium chloride.
M=#mol solute / Liter of solution #mol solute= mass / molar mass #mol solute= 100 / 87 = 1.15 4= 1.15 / L solution 1.15 / 4 = 0.28 _____________________________________________________________ Since we were only given the (M) and grams of LiBr we had to find the mol of solute. To find the #mol of solute we divided the grams (which were given) over the molar mass of LiBr (87). We then got 1.15 . Now we are able to find the Liter of solution . First we plugged in our given values M= 4 & #mol solute= 1.15 . To find liter of solution we simply dived 1.15 over 4 which equals 0.28. --------------- I suppose that lithium bromide is not so soluble to prepare 4 M solutions in water at 20 0C.
To find out the grams of lithium nitrate needed, you need to calculate the molar mass of lithium sulfate (Li2SO4) and lithium nitrate (LiNO3). Then use stoichiometry to determine the amount of lithium nitrate required to produce 250 grams of lithium sulfate. The balanced chemical equation for the reaction would also be needed.