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
Since both barium chloride and barium sulfate contain one mole of barium atoms pert mole of compound, the moles of barium sulfate will be the same, 0.100, when barium has the limiting concentration in the production of the sulfate.
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.
One mole of ammonium nitrate is equal to its molar mass, which is approximately 80.04 grams. This quantity represents Avogadro's number of individual ammonium nitrate molecules.
The chemical formula of barium nitrate is Ba(NO3)2. It is composed of one barium ion (Ba2+) and two nitrate ions (NO3-).
When one mole of sodium nitrate (NaNO3) is added to water, it dissociates into one mole of sodium ions (Na+) and one mole of nitrate ions (NO3-). So, one mole of sodium nitrate produces two moles of solute particles in total when dissolved in water.
Since both barium chloride and barium sulfate contain one mole of barium atoms pert mole of compound, the moles of barium sulfate will be the same, 0.100, when barium has the limiting concentration in the production of the sulfate.
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.
233.43 g On A+ the answer's 233.4 =]
To find the mass of the nitrogen atoms in one mole of cadmium nitrate, calculate the molar mass of cadmium nitrate (Cd(NO3)2) and then multiply it by the number of nitrogen atoms in one mole of cadmium nitrate (2 nitrogen atoms). The molar mass of cadmium nitrate is 236.41 g/mol. Therefore, the mass of the nitrogen atoms in one mole of cadmium nitrate is 28.02 g.
Potassium nitrate is KNO3. There is one potassium per one nitrate. One mole of potassium nitrate contains one mole of nitrate.
One mole of ammonium nitrate is equal to its molar mass, which is approximately 80.04 grams. This quantity represents Avogadro's number of individual ammonium nitrate molecules.
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).
The chemical formula of barium nitrate is Ba(NO3)2. It is composed of one barium ion (Ba2+) and two nitrate ions (NO3-).
One mole of barium nitrate contains six moles of oxygen atoms. One formula contains 6 atoms.
To calculate the mass of nitrogen atoms in one mole of cadmium nitrate, we first need to determine the molar mass of Cd(NO3)2. The molar mass would be the sum of the atomic weights of Cd, N, and 6 O atoms in the formula. After adding these atomic weights, we can find the mass of nitrogen atoms in one mole of cadmium nitrate by multiplying the molar mass by the ratio of nitrogen atoms in the formula.
Using atomic weights for Na = 23 and N=14 and O = 16, one arrives at a mass for 1 mole of NaNO3 of23 + 14 + (3x16) = 85 grams/mole
Correctly it should be written as 'Ba(NO3)2 ' and it is barium nitrate. Notice the use of brackets and the '2' , to indicate that there are two nitrate anions combined to the one barium cation. NB As you gave it, it does not make sense as a chemical formula.