6. How many grams of Nal (149.89 g/mol) are required to produce 92.2 g of Pb12 (461.00 g/mol)?! Pb(NO3)2(aq) + 2Nal...
5. How much heat is produced by the combustion of 5.25 g of acetylene (C2H2, 26.036 g/mol)? 2C2H2+502( 4CO2(@+ 2H20( AH -2602 kJ a) 131 kJ b) 262 kJ c) 496 kJ d) 6830 k e) 13660 kJ 10 6. How many grams of Nal (149.89 g/mol) are required to produce 92.2 g of Pbl2 (461.00 g/mol)? Pb(NO 3)2(aq) + 2NaI (a) Pbl2)+ 2NANO3(ag) a) 567 g b) 284 g c) 60.0 g d) 30.0 g e) 92.9 g
1. A system does 425 kJ of work and loses 125 kJ of heat to the surroundings. What is the change in internal enery of the system? a) 550 kJ c) -300 kJ e) OkJ b) 300 kJ d) - 550 kJ 2. How many grams of NaBr (102.89 g/mol) are contained in 250.0 mL of a 0.200 M solution of sodium bromide a) 50.0 g c) 5.14 g c) 0.0500 g b) 20.6 g d) 0.486 g 3. A...
A student mixes 41.0 mL of 2.84 M Pb(NO3)2(aq) with 20.0 mL of 0.00235 M Nal(aq). How many moles of Pbla(s) precipitate from the resulting solution? Number K,,[Pb12()] – 9810 mol What are the values of [Pb2+), [1]. [NO3-), and [Na*) after the solution has reached equilibrium at 25 °C? Number Number [Pb?*] = Number Number [no; ]- 0 M [nat)-
How many grams of PbCl2 are formed when 35.0 mL of 0.520 M KCl react with Pb(NO3)2? 2KCl(aq) + Pb(NO3)2(aq) → 2KNO3(aq) + PbCl2(s) How many grams of PbCl2 are formed when 35.0 mL of 0.520 M KCl react with Pb(NO3)2? 2KCl(aq) + Pb(NO3)2(aq) → 2KNO3(aq) + PbCl2(s) 10.1 g 25.3 g 5.06 g 15.2 g 2.53 g
Question 3 2 pts 35.5 mL of 0.375 M Pb(NO3)2 (aq) is poured into excess Nal (aq). How many g of Pblz (s) will form? Pb(NO3)2 (aq) + 2 Nal (aq) - --> Pblz (s) + 2 NaNO3(aq) g Pbl2
Reaction 1: Use in question 3 Pb(NO3)2 (aq) + Kl (aq) → KNO, (aq) + Pblz (s) 3. a. When the reaction above is balanced how many moles of lead nitrate are required to react with 2.0 moles of potassium iodide? (1.0 mol Pb(NO3)2) b. How many grams of lead (II) iodide are produced from 5.0 moles of potassium iodide according to the equation given above? (1200 g Pblz)
2NaCl(aq) + Pb(NO3)2(aq)-NaNO3(aq) + PbCl2(s) How many formula units of Pb(NO3)2 would be needed in the above reaction to produce 555kg of PbCl2?
QUESTION 1 In the commercial manufacture of nitric acid, how many moles of NO2(g) are required to produce 250 grams of nitric acid, HNO3(aq) 3NO2(g) + H20(1) - 2HNO3(aq) + NO(g) 69.5 moles 12.7 moles 23.8 moles 5.95 moles 47.6 moles QUESTION 2 The net ionic equation for the reaction of the strong electrolytes lead nitrate and sodium carbonate to form the precipitate lead carbonate is Pb2+(aq) + CO32-(aq) + 2Na+(aq) + 2NO3- (aq) - PCO3(s) + 2NaNO3(aq) Pb2+(aq) 2NO3"(aq)...
student mixes 35.0 mL of 3.18 M Pb(NO3)2(aq) with 20.0 mL of 0.00151 M Nal(aq). How many moles of Pbl2(s) precipitate from the resulting solution? Number 3.02x 10-5mol What are the values of [P NO3, and [Na'] after the solution has reached equilibrium at 25 °C? Number Number Pb2.02 111-114.84 ×10-9 IM Number Number NO4.05
Excess NaHCO3(s) is added to 535 mL of Cu(NO3)2(aq) 0.240 M for the reaction Cu(NO3)2(aq)+2NaHCO3(s)→CuCO3(s)+2NaNO3(aq)+H2O(l)+CO2(g). a)How many grams of NaHCO3(s) the will be consumed? b)How many grams of CuCO3(s) will be produced?