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BaNO32 This is strong electrolyte weak electrolyte or nonelectrolyte?
it is a strong electrolyte because it occurs in strong bases and that's why it is bonded between two different elments
How many moles of oxygen atoms are contained in one mole of BaNO32?
9 moles (there are four oxygen atoms for every mol of BaSO4, so you multiply 2.25 by 4)
What is the chemical product of BaNO32 H3PO4?
Overall reaction including spectator ions: BaCl2(aq)+K2SO4(aq) --> BaSO4(s)+2KCl(aq) Ionic Reaction: Ba+2(aq) + 2Cl-(aq) + 2K+(aq) + SO4-2(aq) --> BaSO4(s) + 2K+(aq) + 2Cl-(aq) Note that the BaSO4(s) is insoluble. Net Ionic: Ba+2(aq) + SO4-2(aq) --> BaSO4(s) The net ionic equation eliminates the ions that just stay in solution and do not contribute to the precipitate of the barium sulfate.
Is BaNO32 soluble in water?
Barium nitrate is soluble in water.
What kind of salt is BaNO32?
This is barium nitrate, an inorganic ionic salt.
BaNO32 This is strong electrolyte weak electrolyte or nonelectrolyte?
it is a strong electrolyte because it occurs in strong bases and that's why it is bonded between two different elments
What is the name of BaNO32?
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.
What is the mass in grams of 0.625 mole bano32?
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.
How many moles of oxygen atoms are contained in one mole of BaNO32?
9 moles (there are four oxygen atoms for every mol of BaSO4, so you multiply 2.25 by 4)
When agno3 reacts with bacl2 agcl and bano32 are formed how many grams of agcl are formed when 10.0g of silver nitrate reacts with 15.0g of bacl2?
To find the amount of AgCl formed, we first need to calculate the limiting reagent. This is done by converting the given masses of AgNO3 and BaCl2 to moles, determining the mole ratio between them, and then the limiting reagent based on the smaller value. Once the limiting reagent is determined, use the mole ratio from the balanced chemical equation to find the moles of AgCl formed and then convert that to grams.
What is the chemical product of BaNO32 H3PO4?
Overall reaction including spectator ions: BaCl2(aq)+K2SO4(aq) --> BaSO4(s)+2KCl(aq) Ionic Reaction: Ba+2(aq) + 2Cl-(aq) + 2K+(aq) + SO4-2(aq) --> BaSO4(s) + 2K+(aq) + 2Cl-(aq) Note that the BaSO4(s) is insoluble. Net Ionic: Ba+2(aq) + SO4-2(aq) --> BaSO4(s) The net ionic equation eliminates the ions that just stay in solution and do not contribute to the precipitate of the barium sulfate.
