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Question 24 (2.57 points) How many milliliters of 0.550 M hydrochloric acid are needed to react...
How many milliliters of 0.550 M hydrochloric acid are needed to react with 15.00 mL of 0.217 M Mg(OH)2? (Answer 11.8 mL, make sure you do the right work) RARE 2HCl(aq) + Mg(OH)2 (aq) + MgCl2(aq) + 2H20(1)
How many milliliters of 0.200 M NH OH are needed to react with 12.0 mL of 0.550 M FeCl3 according to the reaction shown? FeCiglaq) + 3NH2OH(aq) - Fe(OH)3(s) + 3NH4Cl(aq) 68.8 ml 8.25 mL 99.0 mL 33.0 mL 132 mL
How many milliliters of 0.480 M HCl are needed to react with 54.8 g of CaCO3? 2HCl(aq) + CaCO3(s) → CaCl2(aq) + CO2(g) + H2O(l) mL?
7.51 How many milliliters of 6.00 M HCl solution would be needed to react exactly with 20,0 g of pure solid NaOH? HCl(aq) + NaOH(s) — NaCl(aq) + H2O(C)
Zinc reacts with hydrochloric acid according to the reaction equation shown. Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g) How many milliliters of 6.50 M HCl(aq) are required to react with 4.45 g of an ore containing 39.0 % Zn(s) by mass?
Zinc reacts with hydrochloric acid according to the reaction equation shown. Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g) How many milliliters of 4.00 M HCl(aq) are required to react with 7.15 g of an ore containing 43.0 % Zn(s) by mass?
Zinc reacts with hydrochloric acid according to the reaction equation shown. Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g) How many milliliters of 5.50 M HCl(aq) are required to react with 2.95 g of an ore containing 43.0 % Zn(s) by mass?
How many milliliters of a 0,40 M solution of hydrochloric acid (HCI) are necessary to neutralize 33 mL of a 0.20 M barium hydroxide (Ba(OH)2 solution? 16.5 mL 20 ml 66 mL O 33 mL O 10.5 mL
Zinc reacts with hydrochloric acid according to the reaction equation shown. Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g) How many milliliters of 5.00 M HCl(aq) are required to react with 2.65 g of an ore containing 26.0 % Zn(s) by mass? volume: mL
the steps needed to the answer? 9. What volume of a 0.200 M hydrochloric acid (36.46 g/mol) solution is required to completely react with 0.550 g of aluminum hydroxide (78.01 g/mol)? Al(OH)3(s) ЗНСКад) AICI3(aq) + 3H2O(I a. 35.3 mL b) 106 mL c. 0.106 mL d. 11.8 mL