(7.6g LiBr)/(86.84g/mol) x (1molLi/1molBr) = .0875 mol Li or with sig figs, .088
33 grams aluminum (1 mole Al/26.98 grams) = 1.2 moles of aluminum ================
The lowest freezing point is observed for 1 mole of KOH, because its one moles produce 2 moles of ions in solution, 1 mole of cation K+ and 1 mole of anion OH-.
114,426 g carbon dioxide are produced.
The total number of moles of NO produced when 1,0 mole of O2 is completely consumed is 2.
Get moles silver nitrate. 255 grams AgNO3 (1 mole AgNO3/169.91 grams) = 1.5008 moles AgCO3 --------------------------------Now; Molarity = moles of solute/Liters of solution ( 1500 ml = 1.5 Liters ) Molarity = 1.5008 moles AgNO3/1.5 Liters = 1.00 M AgNO3 ---------------------
7,61 g LiBr contain 6,99 g Br.6,99 g Br is 0,087 moles.
33 grams aluminum (1 mole Al/26.98 grams) = 1.2 moles of aluminum ================
Two moles of water are produced.
first determine the number ofmoles dissolved in given solution then .5 moles moles dissolved in 800g. as comparison with 1000g of water, we know 100g of water dissolve only.1 moles of a glucose so we .7moles of glucose dissolve in 800g.
0,028 moles carbonic are obtained.
The lowest freezing point is observed for 1 mole of KOH, because its one moles produce 2 moles of ions in solution, 1 mole of cation K+ and 1 mole of anion OH-.
5.0 grams CO2 (1mol CO2/44.01g) = 0.11 moles CO2
Molarity = moles of solute/Liters of solution Find moles glucose, which is molecular formula------C6H12O6 154 grams C6H12O6 (1 mole C6H12O6/180.156 grams) = 0.8548 moles C6H12O6 Molarity = 0.8548 moles C6H12O6/1 Liter = 0.855 M glucose ---------------------------
The mass of ammonia will be 95,03 g.
1400 grams
114,426 g carbon dioxide are produced.
The total number of moles of NO produced when 1,0 mole of O2 is completely consumed is 2.