The formula of (unhydrated) barium nitrate is Ba(NO3)2. This means, among other things, that each mole contain one mole of barium atoms, two moles of nitrogen atoms, and six moles of oxygen atoms. Thus, the molar mass of this compound is 137.327
+ 2(14.007) + 6(15.999) = 261 grams, to three significant digits. 50 moles will then be fifty times this or 1.3 kilograms, to the justified number of significant digits.
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 ammonium nitrate (NH4NO3) is approximately 80.04 g/mol.
To find the number of moles in 1.25 grams of iron nitrate, we need to know its molar mass. Iron nitrate has a molar mass of approximately 404.0 g/mol. By dividing the mass by the molar mass, we find that 1.25 grams of iron nitrate is roughly 0.0031 moles.
To find the number of moles in 8.0g of ammonium nitrate, you need to divide the given mass by the molar mass of ammonium nitrate (NH4NO3). The molar mass of NH4NO3 is 80.05 g/mol. Therefore, 8.0g ÷ 80.05 g/mol = 0.1 moles of ammonium nitrate.
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.
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
Barium nitrate consists of one barium cation (Ba2+) and two nitrate anions (NO3-). The percent composition can be calculated by finding the molar mass of each element in the compound and dividing by the molar mass of the whole compound. For barium nitrate (Ba(NO3)2), the percent composition of barium is approximately 20.6% (Ba: 137.33 g/mol, Ba(NO3)2: 261.34 g/mol).
233.43 g On A+ the answer's 233.4 =]
The molar mass for anhydrous barium sulfate (BaSO4) is 233.43 g/mol
106
The molar mass of calcium nitrate is 164.1 g/mol. This is equivalent to 164,100 mg/mol.
The molar mass of ammonium nitrate (NH4NO3) is approximately 80.04 g/mol.
To determine the mass of barium in barium phosphate, we need to know the molar ratio of barium to oxygen in the compound. If we assume a 1:3 ratio between barium and oxygen in barium phosphate (Ba3(PO4)2), we can calculate the molar mass of barium and oxygen in the compound. With this information, we can find the mass of barium in the sample by subtracting the mass of oxygen from the total mass.
get the volume of the chemical and the mass and divide the mass by it's volume
The molar mass of BaSO4 (Barium sulfate) can be calculated by adding the molar mass of each element present in the formula: Ba (barium) has a molar mass of 137.33 g/mol, S (sulfur) has a molar mass of 32.06 g/mol, and O (oxygen) has a molar mass of 16.00 g/mol. Adding these together gives a molar mass of 137.33 + 32.06 + (4 * 16.00) = 233.37 g/mol for BaSO4.
169.8731 g/mol
To determine the number of moles in 1 g of zinc nitrate, you need to know the molar mass of zinc nitrate. Zinc nitrate has a molar mass of 189.36 g/mol. By dividing 1 g by the molar mass, you can calculate that there are approximately 0.0053 moles of zinc nitrate in 1 g.