Mainly, you need to find out how many liters there are in a mole. This varies from one substance to another.
2.5 g per mL = 2,500 g per Litre.
To calculate the grams of LiCl needed to make a 2.0 L solution at a concentration of 0.65 M, use the formula: [ \text{grams of solute} = \text{molarity} \times \text{volume (L)} \times \text{molar mass} ] The molar mass of LiCl is approximately 42.39 g/mol. Thus: [ \text{grams of LiCl} = 0.65 , \text{mol/L} \times 2.0 , \text{L} \times 42.39 , \text{g/mol} \approx 55.85 , \text{g} ] Therefore, approximately 55.85 grams of LiCl are required.
You begin with 1.63 g of MgCl2•xH2O. You end up with 0.762 g MgCl2. That means you lost 1.63 - 0.762 = 0.87 g of H2O. 0.762 g MgCl2 / 95.21 g (molar mass) = .00800 mol MgCl2 0.87 g H2O / 18.02 g (molar mass) = .048 mol H2O x = mol solute / mol solvent x = .048 mol H2O / .00800 mol MgCl2 x = 6 MgCl2•6H2O
To find the Gibbs free energy change (ΔG) at 500 K, we can use the equation ΔG = ΔH - TΔS. Given that ΔH = -27 kJ/mol and ΔS = 0.09 kJ/(mol·K), we first convert the temperature to Kelvin (which is already given as 500 K). Then, substituting the values: ΔG = -27 kJ/mol - (500 K × 0.09 kJ/(mol·K)) = -27 kJ/mol - 45 kJ/mol = -72 kJ/mol. Thus, the value of G at 500 K is -72 kJ/mol.
To determine how many grams of NH₃ can be produced from 3.13 mol of N₂, we start with the balanced chemical equation: N₂ + 3H₂ → 2NH₃. From the equation, 1 mole of N₂ produces 2 moles of NH₃. Therefore, 3.13 moles of N₂ will produce 3.13 × 2 = 6.26 moles of NH₃. To convert moles of NH₃ to grams, we use its molar mass (approximately 17.03 g/mol): 6.26 mol × 17.03 g/mol = 106.36 grams of NH₃.
To find the molarity of the Na3PO4 solution, we first need to calculate the molar mass of Na3PO4: Na: 22.99 g/mol P: 30.97 g/mol O: 16.00 g/mol Total molar mass = 22.99 * 3 + 30.97 + 16.00 * 4 = 163.94 g/mol Next, we find the concentration in mol/L: 0.142 M Na3PO4 = 0.142 mol/L Then, we convert the density from g/mL to g/L: 1.015 g/mL * 1000 mL/L = 1015 g/L Finally, we calculate the molarity using the formula: molarity = (density / molar mass) * concentration molarity = (1015 g/L / 163.94 g/mol) * 0.142 mol/L ≈ 0.88 M
To calculate concentration from molarity, you can use the formula: concentration (in g/L) molarity (in mol/L) x molar mass (in g/mol). This formula helps you convert the molarity of a solution into its concentration in grams per liter.
4.25 grams. .050 M = .050 mol/1 L 5.0 L x .050 mol/L (cancel out L to get mol as a unit)= .25 mol Atomic mass of Ammonia (NH3)= 17 g/mol .25 mol x 17 g/mol (cancel out mol to get g as a unit)= 4.25 g
Molarity
To calculate the molarity, first convert the mass of LiCl to moles using its molar mass (6.94 g/mol for Li, 35.45 g/mol for Cl). Then, divide the moles of LiCl by the volume in liters (930 mL = 0.93 L) to get the molarity. The molarity of the solution would be around 5.2 mol/L.
mol/L
To calculate the mass of iodine needed to prepare a 0.200 N solution in 50 mL, you can use the formula: Mass (g) = molarity (mol/L) * volume (L) * molar mass (g/mol). First, convert 50 mL to L (50 mL = 0.050 L). Then, substitute the values: Mass (g) = 0.200 mol/L * 0.050 L * 253.8 g/mol = 2.538 g of iodine. Therefore, you would weigh out 2.538 grams of iodine.
To convert grams (g) to millimoles per liter (mmol/L) for a substance, you need to know the molar mass of the substance. Then you can use the formula: Concentration in mmol/L = (mass in g) / (molar mass in g/mol) * 1000.
The conversion factor to convert daltons (Da) to grams per mole (g/mol) is 1 Da 1.66054 x 10-24 g/mol.
Examples are: mol/L, g/L, g/1oo mL.
The molar mass of NaCl is 58.44 g/mol. To calculate the molarity, first convert grams to moles (58 g / 58.44 g/mol = 0.993 mol). Then divide moles by liters of solution (0.993 mol / 1.0 L = 0.993 M). Therefore, the molarity of the solution is 0.993 M.
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