Approx. 516 g at 0 0C.
The chemical reaction is:Li2CO3 + Ca(OH)2 = 2 LiOH + CaCO3Molar maass of LiOH is 23,95 g; molar mass of calcium hydroxide is 74,093 g.74,093 g Ca(OH)2----------------2.23,95 g LiOH25 g Ca(OH)2----------------------xx = (25 x 2 x 23,95)/74,093 = 16,2 g LiOH
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 grams of lithium hydroxide present, you need to use its molar mass. The molar mass of LiOH is approximately 23.95 g/mol. Therefore, 3 moles of LiOH would be: 3 moles x 23.95 g/mol = 71.85 grams of lithium hydroxide.
First, determine the limiting reactant by calculating the number of moles for each reactant (moles = concentration x volume). In this case, LiOH is the limiting reactant since it produces 2 moles of water per 1 mole of LiOH. Therefore, 0.006 moles of LiOH will produce 0.012 moles of water.
There are 700 grams in 700 grams.
The answer is 445,6 g carbon dioxide.
There are 5800 grams in 5.8 kg since 1 kg is equal to 1000 grams.
1000 Grams per kilogram, so 5800:1000 = 5.8 Kg.
To find the grams of LiOH in the solution, we need to use the percentage concentration of NaOH to calculate the molarity of the solution first. Then, we can use stoichiometry to convert the moles of NaOH to LiOH since they are in a 1:1 ratio. Finally, we can convert the moles of LiOH to grams using its molar mass.
Two moles of LiOH may absorbe one mole of CO2 that means 47.9g LiOH may absorbe 44.01g CO2 so 1 g may absorbe 44.01 divided by 47.9 = 0.919 g. 2LiOH + CO2 = Li2CO3 + H2O
The chemical reaction is:Li2CO3 + Ca(OH)2 = 2 LiOH + CaCO3Molar maass of LiOH is 23,95 g; molar mass of calcium hydroxide is 74,093 g.74,093 g Ca(OH)2----------------2.23,95 g LiOH25 g Ca(OH)2----------------------xx = (25 x 2 x 23,95)/74,093 = 16,2 g LiOH
5800 ' = 1.09848 mi.
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)
There are 3 feet in one yard. Therefore, 5800 feet is equal to 5800/3 = 1933.3 recurring (that is, 1933.3333...) yards.
To determine the grams of lithium hydroxide present, you need to use its molar mass. The molar mass of LiOH is approximately 23.95 g/mol. Therefore, 3 moles of LiOH would be: 3 moles x 23.95 g/mol = 71.85 grams of lithium hydroxide.
The balanced chemical equation for the reaction of lithium hydroxide with carbon dioxide is 2 LiOH + CO2 -> Li2CO3 + H2O. The mole ratio of LiOH to CO2 is 2:1, meaning that 40 moles of LiOH are required to react with 20 moles of CO2.
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