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a 10 ml solution is 0.50 m silver nitrate. what minimum volume (in milliliters)in 0.10m sodium acetate is required to ensure precipitation of silver acetate given that ksp(AgCH3COO)= 4.0x10^-3.
Minimum volume=367.4mL
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When sodium chloride reacts with a silver nitrate silver chloride precipitates what mass of a AgCl it’s produce from 75.0 g of AgNO<sub>3</sub>
NaCl+AgNO3→AgCl+NaNO3Mole ratio = 1:1Molar Mass of AgNO3 = 169.87= 75/169.87= 0.442 molMolar Mass of AgCl = 143.32= 143.32 × 0.442= 63.278g
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Reaction between lead(ii) nitrate and sodium iodide to produce lead(ii) iodide and sodium nitrate
Pb(NO3)2 (aq) + 2 NaI (aq) → PbI2 (s) + 2 NaNO3 (aq)2Na(+) +2I(-) + Pb(2+) + 2NO3(-) = 2Na(+) +PbI2(s) + 2NO3(-)2I(-) + Pb(2+) = PbI2 (s)Lead iodide appears as a yellow crystalline solid. Insoluble in water.
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If 5.21 g of silver nitrate react with sodium phosphate what mass of silver phosphate is produced?
Question: If 5.21 g of silver nitrate react with sodium phosphate what mass of silver phosphate is produced? m(AgNO3) = 5.21 g m(Ag3PO4) -? Solution: 3AgNO3 + Na3PO4 = Ag3PO4...
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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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explain how you would obtain solid sodium carbonate from a mixture of lead (II) carbonate and sodium carbonate.
As both sodium carbonate and lead carbonate both are solid but their solubility are different due to presence of cations. As in sodium carbonate sodium is the cation which is...
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Silver carbonate is most soluble in_________. *
A.SODIUM CARBONATE
B.NITRIC ACID
C.SILVER NITRATE
Silver carbonate is most soluble inA.SODIUM CARBONATE
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<ol><li>A student has a piece of silver. The density of silver is 14.09g/cm3. If the mass of the piece of silver is 89.05g, what is the volume, in cm3, of the piece of silver? </li></ol>
Accordng to the equation, the volume can be determined as following:V = m/dwhere m - mass, d - density, V - volume.From here:V = 89.05 g / 14.09g/cm3 = 6.32...
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Will a precipitate form if 120 mL of 0.045 M silver nitrate is mixed with 180 mL of 0.052 M sodium
chromate? (silver chromate has Ksp = 1.9 x10^–12) Calculate [Ag+] and [CrO4^2–] at equilibrium.
2Ag+ (aq) + CrO42- (aq) ↔ Ag2CrO4 (s)mol Ag+ (aq) = (0.120 L)(0.045 mol/L) = 0.0054 mol Ag+[Ag+] = 0.0054 mol/(0.120 L + 0.180 L) = 0.0054 mol / 0.300...
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