Need the ionic equation and net ionic equations 1. AgNO3(aq) +KCl(aq) ----> KNO3 + AgCl 2. CuCl2(aq) + Ba(NO3)2(aq) -----> Cu(NO3)2 + BaCl2 Thank you so much for helping me!

Md Ariful Islam

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1. AgNO3(aq) +KCl(aq) ----> KNO3(aq) + AgCl (s) Ag+ + NO3- +K+ + Cl- ----> K+ + NO3- + AgCl(s) Ag+ + Cl- ----> AgCl(s) 2. CuCl2(aq) + Ba(NO3)2(aq) -----> Cu(NO3)2 (aq) + BaCl2 (aq) Cu2+ + 2Cl- + Ba2+ + 2NO3- -----> Cu2+ + 2NO3- + Ba2+ + 2Cl- No reaction
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What mass of precipitate (in g) is formed when 58.2 mL of 0.500 M FeCl₃ reacts with excess AgNO₃ in the following chemical reaction? FeCl₃(aq) + 3 AgNO₃(aq) → 3 AgCl(s) + Fe(NO₃)₃(aq)

Mass of precipitate =2.37g

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Fill in the appropriate unit conversions. (Report the molar mass of KCl using three sig. figs.) ? mol/L​ = 5.00 g KCl/100. mL​×1 mol KCl/ KCl​×/1 L ​

Molar mass of KCl = 74.6g/mol

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A quality analyst mixes 3ml of 0.255M of Bacl2 with excess Agno3 causing Agcl to precipitate and subsequently isolate 2.01g of Agcl.calculate the yield of the reaction

Molar mass of BaCl2 = 208.23 g/mol0.255 × 208.23 = 53.0992/53.099 × 100= 3.7665%

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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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need the ionic and net ionic of the equation 1. CuSO4(aq) + CuCl2(aq)----->CuCl2 + CuSO4 2. CuCl2(aq) + KCl(aq)----->CuCl2 + KCl

1. CuSO4(aq) + CuCl2(aq) -----> no reaction Cu2+(aq) + SO42-(aq) + Cu2+(aq) + 2Cl-(aq) -----> no reaction 2. CuCl2(aq) + KCl(aq)-----> no reaction Cu2+(aq) + 2Cl-(aq) + K+(aq) + Cl-(aq)...

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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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in the reaction KNO3(aq)+PbSO4(aq)-->Pb(NO3)2(aq)+K2SO4(aq), what is the oxidizing agent? a. KNO3(aq) b. PbSO4(aq) c. Pb(NO3)2 d.K2SO4(aq) e. there is no oxidizing agent

e. there is no oxidizing agent

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When 50cm of a saturated solution of knO3 at 25°c was evaporated to dryness ,10g of dry salt was obtained what is the solubility of knO3 at 25°c (knO3=101)

1mol =100cm3? = 50cm350/1000 =0.05 mol1mol = 101 g0.05mol =[ 0.05][101]=5.05g{[10][100]}/[5.05]=198g/cm3

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3.38 g of AgNO3 react with BaCl2 to produce 1.43 g of AgCl according to the following balanced equation: 2AgNO3 + BaCl2 → 2 AgCl + Ba(NO3)2 What is the percent yield of AgCl in this reaction?

According to the equation, moles AgNO3 = moles AgCl.So the theoretical yield of this reaction can be calculated as: mtheor(AgCl) = (m(AgNO3) / M(AgNO3)) x M (AgCl) = (3.38g /...

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50 ml of HCl solution (1 mol/L) were added to 53 g of AgNO3 solution (0.22 w/w%). How many grams of AgCl were produced? The molecular masses are 169.9 g/mol (AgNO3) and 143.3 g/mol (AgCl).

HCl + AgNO3----------- AgCl + HNO3To get our limiting reagent that will be used for the reaction:Using n=CV for HclC=1mol/L V=50/1000= 0.05Ln=1×0.05=0.05molesTherefore using a mole-mass relationship If 1mole of HCl...

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