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0.1868 moles
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How many moles are contained in 22.23 g KBr?
To find the number of moles in 22.23 g of KBr, we need to divide the given mass by the molar mass of KBr. The molar mass of KBr is 119 g/mol. Therefore, 22.23 g of KBr is equal to 0.187 moles.
How many moles of kbr are present in 25ml of a 1.5 M solution?
To find the number of moles of KBr in the solution, first calculate the number of moles of KBr in the 25 mL solution using the given concentration and volume. $$moles = concentration \times volume$$ Then, multiply the moles by the molecular weight of KBr to get the mass of KBr in the solution if needed.
How many moles of KBr will be produced from 4.79 moles of BaBr2?
14.17 mol BaBr2 has 2*14.17 mol Br in it, so 28.34 mol KBr can be produced (also 28.34 mol K is needed)
How many moles of KBr are present in 110mL of a 0.290M solution?
To find the number of moles of KBr in the solution, first convert the volume to liters (110mL = 0.110L). Then, use the formula: moles = molarity x volume in liters. So, moles of KBr = 0.290 mol/L x 0.110 L = 0.032 moles of KBr.
How many moles of KBr will be produced from 7 moles of BaBr2?
None, unless there is metallic potassium in the reaction mixture. Assuming excess potassium metal is present then 14 moles of KBr can be produced. 7BaBr2 + excess potassium -----> 14KBr + 7 Ba
How many moles of KBr are present in 190mL of a 0.265M solution?
The answer is 0,0509 mole.
How many grams of KBr are present in 300.0 mL of a 1.25 M solution?
To find the grams of KBr in the solution, first calculate the moles of KBr present by using the molarity formula: moles = Molarity x Volume (L). Then, convert moles of KBr to grams using its molar mass. For KBr, the molar mass is approximately 119 g/mol. Finally, perform the calculation to find the grams present in the solution.
How many moles of k are needed to form 3.3 moles of KBr?
357
How many moles are contained in 245g of potassium bromide?
To find the number of moles in 245g of potassium bromide, first calculate the molar mass of KBr by adding the atomic masses of potassium (39.10 g/mol) and bromine (79.90 g/mol). The molar mass of KBr is 119.00 g/mol. Then, divide the given mass by the molar mass: 245g / 119.00 g/mol = 2.06 moles of potassium bromide.
How many moles of potassium bromide can be produced from the reaction of 2.92 moles of potassium with 1.78 moles of bromine gas?
First write a balanced chemical equation: 2K + Br2 ---> 2KBR Find the limiting reactant by using the moles of each element and determining which one gives you the smallest number of moles of potassium bromide. 2.92 mol K (2 mol KBr/2 mol K)= 2.92 mol KBr 1.78 mol Br2 (2 mol KBR/1 mol Br2)=3.56 mol KBr potassium is your limiting reactant so the max. number of moles of KBr that can be produced is 2.92 mol of KBr
How many moles of KBr will be produced from 14.17 moles of BaBr2?
14.17 mol BaBr2 has 2*14.17 mol Br in it, so 28.34 mol KBr can be produced (also 28.34 mol K is needed)
How many grams of KBr are required to make 750.0 mL of solution that is 0.0552M molar mass of KBr 119.00gmol?
Multiply the molarity (M, which is in mol/L) with the volume (in L) to get the number of moles needed. Then multiply the result with the molar mass. If you look at the units they will cancel to give an answer in grams. (mol/L)*(L)=mol, (mol)*(g/mol)=g So for the numerical answer you get (0.0552 mol/L)*(0.750 L)*(119.00 g/mol)= 4.93 g KBr
