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the chemical formula for silver nitrate (AgNO3). Determine the name of each element in the compound, the number of atoms of each element, and the total number of atoms in the chemical formula.
Ag- Silver 1 atomN - Nitrogen 1 atomO - Oxygen 3 atomsTotal atoms = 5 atoms
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An aqueous solution has 18% of solute. By chemical analysis, it was known that it has 10 mol% solute.
What is the molecular weight of the solute?
m = n*M, m - weight, n - quantity of substance, M - molecular weight mx /18 = mH2O / 82nx = nH2O / 9nx *Mx = nH2O * MH2O*18/82Mx...
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6) AgNO3 + K3PO4 Ag3PO4+ KNO3 Mass of AgNO3: 25.0g
Experimental Mass of K3PO4: 9.53g Experimental Yield of Ag3PO4: 18.62g Experiment Yield of KNO3: 13.24g
Molar mass of AgNO3: 169,872 g/mol Molar mass of K3PO4: 212.264 g/mol Molar mass of Ag3PO4: 418.574 g/mol Molar mass of KNO3: 101.102 g/mol. n = m/M n (AgNO3) =...
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How many grams of copper are required to react with 45.0 grams of silver nitrate? Cu+ 2 AgNO3= Cu(NO3)2 + 2 Ag
Solution:The balanced chemical equation:Cu + 2AgNO3= Cu(NO3)2 + 2AgAccording to the equation: n(Cu) = n(AgNO3)/2Moles of AgNO3 = Mass of AgNO3 / Molar mass of AgNO3The molar mass of AgNO3...
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What is the balanced equation to a solution of nitrate is combined with a solution of sodium carbonate to produce sodium nitrate solution and solid silver carbonate
Na2CO3 + 2AgNO3 = Ag2CO3 + 2NaNO3
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If 25 mL of a 0.45 M AgNO3 solution is added to 55 mL of a 1.2M NaCl solution. What mass of AgCl will be precipitated.
AgNO3 (s) + NaCl(aq) ==>> AgCl(s) + NaNO3 (aq)
Calculate moles of both: n=C×V n(AgNO_3 )=0.45 mol/L×0.025 L=0.01125 mol n(NaCl)=1.2 mol/L×0.055 L=0.066 mol NaCl is in excess, AgNO3 is limiting reagent. Thus, n(AgCl)=n(AgNO_3 )=0.01125 mol As we know: n=m/MW...
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What is the concentration of NaCl in a solution, if titration of 15.00mL of the solution with 0.257 M AgNO3 needs 20.22 mL of the AgNO3 solution to reach the end point?
Answer: 0.346 M
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A 0.4230 g sample of impure sodium nitrate was heated, converting all the sodium nitrate to .1235 g of sodium nitrate and oxygen gas. Determine the percent of sodium nitrate in the original sample.
Answer:2HNO3 = 2HNO2 + O2m= 85 x 0.1235 / 69= 0.1521W%= 0.1521 / 0.4230 = 36%
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how many moles of silver nitrate ( AgNO3) are produced from 10.3 moles of nitric acid (HNO3)?
5.15 moles
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the number of atoms present in 215 g of silver nitrate, AgNO3
N=(m(AgNO3)/Mr(AgNO3) )*Na=(215/170)*6.02*10^23=7.613*10^23
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