Molar Mass: ~137.33 g/mol
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Periodic Tables usually come with molar mass, atomic number, symbol, and element name. They're quite helpful.
The molar mass of barium (Ba) is approximately 137.3 g/mol. So, the mass of 3 moles of barium would be 3 moles x 137.3 g/mol = 411.9 grams.
To find the grams of Ba(CH3COO)2 in the solution, calculate the mass of Ba(CH3COO)2 using the percentage data (25.7%) provided. Mass of Ba(CH3COO)2 in 237 g solution = 237 g x 25.7% = 60.99 g Therefore, there are 60.99 grams of Ba(CH3COO)2 in the solution.
The molar mass of SO2 (sulfur dioxide) is approximately 64.06 g/mol. To find the mass of 0.75 moles of SO2, you would multiply the number of moles (0.75) by the molar mass (64.06 g/mol) to get the mass of 0.75 moles of SO2, which is 48.045 grams.
To find the number of moles in 2400 grams of Ba(OH)2, divide the given mass by the molar mass of Ba(OH)2. The molar mass of Ba(OH)2 is 171.34 g/mol. So, 2400 g / 171.34 g/mol = 14.00 mol of Ba(OH)2.
5.25 grams Ba(OH)2 (1mol Ba(OH)2/171.316 grams Ba(OH)2 ) = 0.0306 moles Ba(OH)2
To find the mass of 0.625 moles of Ba(NO3)2 (barium nitrate), first calculate its molar mass. The molar mass of Ba(NO3)2 is approximately 137.33 g/mol (for Ba) + 2 × (14.01 g/mol for N) + 6 × (16.00 g/mol for O), totaling about 261.34 g/mol. Thus, the mass of 0.625 moles is 0.625 moles × 261.34 g/mol ≈ 163.35 grams.
The molar mass of barium (Ba) is approximately 137.3 g/mol. So, the mass of 3 moles of barium would be 3 moles x 137.3 g/mol = 411.9 grams.
To determine how many barium (Ba) atoms would have the same mass as the calculated atomic mass of the element, you can use Avogadro's number, which is approximately (6.022 \times 10^{23}) atoms/mole. The atomic mass of barium is about 137.33 amu, meaning one mole of barium atoms weighs approximately 137.33 grams. Therefore, to find the number of atoms that would have a mass equal to the atomic mass of barium (in grams), you would convert 137.33 grams to moles, which would result in approximately (1) mole or (6.022 \times 10^{23}) barium atoms.
To determine the number of moles present, we first need to find the molar mass of barium (Ba), which is approximately 137.33 g/mol. Next, we use the formula n = m/M, where n is the number of moles, m is the mass of the sample (22.3 grams), and M is the molar mass (137.33 g/mol). By substituting these values, we find that there are approximately 0.162 moles of barium in the sample.
Barium siliconhexafluoride. 279.39 grams BaSiF6 = 1 mole of BaSiF6/6.022 X 10^23 = 1.66 X 10^-24, which is the mole ratio of one molecule of BaSiF6. So, ... (279.39 grams) * (1.66 X 10^-24) = 4.64 X 10^-22 grams in one molecule of BaSiF6
To find the grams of Ba(CH3COO)2 in the solution, calculate the mass of Ba(CH3COO)2 using the percentage data (25.7%) provided. Mass of Ba(CH3COO)2 in 237 g solution = 237 g x 25.7% = 60.99 g Therefore, there are 60.99 grams of Ba(CH3COO)2 in the solution.
First, we need to find the formula for barium nitrate. Barium's charge is 2+, and nitrate is a 1-. If you criss-cross charges, the resulting formula is Ba(NO3)2. Now that we have the formula, we need to find the atomic mass. Barium has an atomic mass of 137. Nitrate is composed of nitrogen (14.01g) and oxygen (16.00*3 = 48). The atomic mass of nitrate is 62.01. Since there are two NO3 molecules, we multiply by 2 to get 124.02. Finally, we add the barium for a grand total of 261.02. Finally, we set-up our conversion ratios: 432 g Ba(NO3)2/1 * 1 mol/261.02g = 1.66 mols
The molar mass of SO2 (sulfur dioxide) is approximately 64.06 g/mol. To find the mass of 0.75 moles of SO2, you would multiply the number of moles (0.75) by the molar mass (64.06 g/mol) to get the mass of 0.75 moles of SO2, which is 48.045 grams.
To find the number of moles in 2400 grams of Ba(OH)2, divide the given mass by the molar mass of Ba(OH)2. The molar mass of Ba(OH)2 is 171.34 g/mol. So, 2400 g / 171.34 g/mol = 14.00 mol of Ba(OH)2.
5.25 grams Ba(OH)2 (1mol Ba(OH)2/171.316 grams Ba(OH)2 ) = 0.0306 moles Ba(OH)2
The molar mass of BaSO4 is 233.4 g/mol. Therefore, 0.0891 g of BaSO4 corresponds to 0.000382 moles of BaSO4. Since BaSO4 contains 1 mole of Ba for every 1 mole of BaSO4, the original sample contained 0.000382 moles of Ba. The percentage of barium in the compound is calculated as (mass of Ba / total mass of the compound) * 100, which equals approximately 0.143%.
To find the number of moles, divide the given mass of Ba(OH)2 (2400g) by its molar mass. The molar mass of Ba(OH)2 is calculated as 137.33 (Ba) + 216.00 (O) + 21.01 (H), resulting in approximately 171.35 g/mol. Dividing 2400g by 171.35 g/mol gives approximately 14.01 moles of Ba(OH)2.