1.- What mass of lithium phosphate is needed to prepare 575 mLof a solution havi

Question

1.- What mass of lithium phosphate is needed to prepare 575 mLof a solution having a lithium ion concentration of 0.250 M? a) 15.5 g   b) 23.2 g c) 44.4 g d) 70.0g e) 5.55 g 2.- 75.0 mL of 0.355 M hydrochloric acid is added to 25.0 mLof 0.298 M MgCl2 . What is the concentration of chlorideion in the final solution? a) 0.415 M b) 0.255M c)0.737M d) 0.949M   e) 0.0255 M 3.- How many grams of water are formed when 17.5 g of oxygenreacts with nitrogen?(hint: you need to write a balancedeqn.)     a) 39.4   b) 9.85 c)19.7 d) 156 e) 78.2 1.- What mass of lithium phosphate is needed to prepare 575 mLof a solution having a lithium ion concentration of 0.250 M? a) 15.5 g   b) 23.2 g c) 44.4 g d) 70.0g e) 5.55 g 2.- 75.0 mL of 0.355 M hydrochloric acid is added to 25.0 mLof 0.298 M MgCl2 . What is the concentration of chlorideion in the final solution? a) 0.415 M b) 0.255M c)0.737M d) 0.949M   e) 0.0255 M 3.- How many grams of water are formed when 17.5 g of oxygenreacts with nitrogen?(hint: you need to write a balancedeqn.)     a) 39.4   b) 9.85 c)19.7 d) 156 e) 78.2

Explanation / Answer

1) Li3PO4 0.575 L *(0.250 mol Li+/1 L) *(1mol Li3PO4 /3molLi+)*(115.80 g/1mol) = 5.55 grams 2) 0.075 L *(0.355 mol Cl-/1L) = 0.026625 mol Cl- 0.025 L *(0.298 mol MgCl2/1L) *(2mol Cl-/1mol MgCl2) = 0.0149 molesCl- Total moles Cl- = 0.026625 + 0.0149 = 0.041525 moles Cl- Final Volume = 75 mL + 25 mL = 100 mL = 0.1 L [Cl-] = 0.041525 moles /0.1 L = 0.415 M   Pick a)

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