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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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Balance and classify the following reaction
AgNO3(aq) + NaBr(aq) → AgBr(s) + NaNO3(aq)
AgNO3(aq) + NaBr(aq) → AgBr(s) + NaNO3(aq)Classification the following reaction is Double-replacement reactions.
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From an initial mix of: 2 mol Cl2, 1 mol Br2, 0.5 mol I2, 0.5 mol Cl−, 1 mol Br−, 2 mol I−; the total quantity of I2 present after all possible reactions occur is
0.5 mol I2 and 0.5 mol Cl- will not react with anything; 2 mol I- will react with one mol of Cl2 producing one mol of I2: 2I- + Cl2...
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The solubility product (Ksp) or AgBr(s) is 2.0 x 〖10〗^(-13). Calculate the concentration of
soluble silver ion in a 0.05 M NaBr solution that was saturated with AgBr(s)?
AgBr(s) → Ag+(aq) + Br-(aq) Initial … 0 0.05M Change … +1xM +xM Equilibrium … xM x+0.05 M Ksp = [Ag+][Br-] = x(x + 0.05) = x·0.05 = 2·10^-13 x...
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1.At 25 ℃, 0.831 mg AgBr dissolves in 9.25 L of water. What is the equilibrium constant for the reaction below?
AgBr(s) ⇆ Ag+(aq) + Br–(aq)
Answer: Ksp = [Ag+][Br-] 0.831 mg = 0.000831 g n = m/M 0.000831 g / 187.78 g/mole = 4.425×10^-6 mole с = n/V c = 4.425×10^-6 / 9.25 = 4.78×10^-7...
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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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How many grams of silver bromide, AgBr, are produced if 15.0 mL of 0.350 M CaBr2(aq) completely react given the reaction:
2 AgNO3(aq) + CaBr2(aq) → Ca(NO3)2(aq) + 2 AgBr(s)
n(CaBr2)= V(CaBr2)×c(CaBr2)=0.015×0.35=0.00525 moln(AgBr)=2×n(CaBr2)=2×0.00525=0.0105 molm(AgBr)=n(AgBr)×M(AgBr)=0.0105×188=1.974 g.
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When the following equation is balanced, what will be the coefficient of AgBr(aq)? Give your answer as a number.
Fe2(CO3)3(aq) + AgBr(s) ⟶ FeBr3(aq) + Ag2CO3(s)
6
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How many grams of AgBr will form when 45.0 mL of 0.842 M CaBr2 react?
CaBr2(aq) + 2 AgNO3(aq) → 2 AgBr(s) + Ca(NO3)2(aq)
Solution:Molar mass of AgBr = 187.77 g/molmass of AgBr = M of CaBr2× V of CaBr2 / 1000ml × n of AgBr × molar mass of AgBr = 0.842molCaBr2× 45mlCaBr2/1000ml×2×187.77g/molAgBr...
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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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