Ba(NO3)2
The total mass for this molecule is 261g
so what is 261g/132g this is 1.97moles
Ba(NO3)2 and Al(OH)3
To find the mass in grams of 4.5 x 10²² molecules of barium nitrate, Ba(NO₂)₂, first determine the molar mass of Ba(NO₂)₂. The molar mass is approximately 199.34 g/mol. Next, convert the number of molecules to moles using Avogadro's number (6.022 x 10²³ molecules/mol): [ \text{moles} = \frac{4.5 \times 10^{22}}{6.022 \times 10^{23}} \approx 0.0747 \text{ moles}. ] Finally, multiply the number of moles by the molar mass to find the mass: [ \text{mass} = 0.0747 \text{ moles} \times 199.34 \text{ g/mol} \approx 14.89 \text{ grams}. ]
[(aa + bb) + (ab+ba)(aa+bb)*(ab+ba)]*[a + (ab+ba)(aa+bb)*b]
If they were lauhing after it yea but if he/she was serious looking about it no... if you deffinitally think its not just call the numner and ask if ba ba ba there?
when you multiply a positive number by a negative number it is a negative. if u multiply a negative by a negative then it is positive
9 moles (there are four oxygen atoms for every mol of BaSO4, so you multiply 2.25 by 4)
Ba(NO3)2 Molarity = moles of solute/Liters of solution ( 100 ml = 0.100 liters ) 0.10 M Ba(NO3)2 = moles Ba(NO3)2/0.100 liters = 0.01 moles Ba(NO3)2 (261.32 grams/1 mole Ba(NO3)2) = 2.6 grams of Ba(NO3)2 needs to be put into that 100 milliliters of solution.
Ba(NO3)2 contains a total of 6 oxygen atoms - there are 3 oxygen atoms in each nitrate (NO3) group, and there are 2 nitrate groups in Ba(NO3)2.
3,75 moles barium chloride
Ba3(PO4) is an incorrect formula. Barium phosphate has the formula 'Ba3(PO4)2 '. So write down the BALANCED reaction eq'n. 2Na3PO4 + 3Ba(NO3)2 = 6NaNO3 + Ba3(PO4)2 The molar ratios are 2:3 :: 6:1 We have 3.25 mol(Na3PO4) equivalent to '2' & 7.03 mol(Ba(NO3)2 ) equivalent to '3' We now must find the limiting reactant. So 3..25 / 2 = 1.625 & 7.03/3 = 2.34333.... We only need 1.625 moles(Na3PO4) to react with 7.03 moles (Ba(NO3)2). So at 3.25 moles the sodium phosphate is in excess. Therefore 7.03 moles barium nitrate is the limiting reactant. By equivalence 3:1 ;; 7.03 : x Hence x /1 = 7.03 / 3 x = 7.03 x 1 / 3 x = 2.34333... moles(Ba(NO3)2) is produced.
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
One mole of barium nitrate contains six moles of oxygen atoms. One formula contains 6 atoms.
The chemical formula for Barium Nitrite is Ba(NO2)2. Ba = Barium N = Nitrogen O = Oxygen Its molar mass is 229.338 g/mol.
Ba(NO3)2(aq) + Na2SO4(aq) ==> BaSO4(s) + 2NaNO3(aq).
The chemical formula for Barium ion (Ba2+) is simply Ba2+ and the chemical formula for Nitrate ion (NO3-) is NO3−. To represent the compound formed by Barium ion and Nitrate ion, write Ba(NO3)2.
471,6 g of anhydrous Ba(OH)2 is equivalent to 2,75 moles.
Ba(NO3)2 is the chemical formula for barium nitrate, a compound composed of one barium ion (Ba^2+) and two nitrate ions (NO3^-). It is commonly used in fireworks to produce a green flame.