What is the of a 3.5 times 10^2 mL solution that contains 5.43 g of KCI? 15.5 M

Question

What is the of a 3.5 times 10^2 mL solution that contains 5.43 g of KCI? 15.5 M 4.81 M 2.08 M 0.208 M What volume of 10.2 M HCI is required to prepare 50.0 mL of a 0.500 M HCI solution? 49.0 mL 4, 90 mL. 2.45 mL 0.510 mL What is the concentration of a 15.0 ml phosphoric acid solution that requires 10, 0 mL of 0.200 M NaOH to react completely? H_3PO_4(aq) + 3Na OH (aq) rightarrow N PO_4 (aq) + 3H_2 O(I) 10.0 M 0.400 M 0.100 M 0.0444 M For the reaction shown in the diagram how many moles of oxygen, O_2, are required to produce 1. 0 mole of the product? 4.0 2.0 1.5 1.0

Explanation / Answer

45) molarity = moles / Ltr of solution

moles = 5.43 g / 74.55 g/mol = 0.0728 moles

3.50 x 102 mL = 350 mL = 0.350 Ltr

molarity = 0.0728 moles / 0.350 Ltr

molarity = 0.208 M (option D)

46) V1M1 = V2M2

V2 = 50 mL * 0.500 M / 10.2 M

V2 = 2.45 ( option C)

47)

M of H3PO4 = 3 ( 10 mL * 0.200 M) / 15 mL

M of H3PO4 = 0.4 M option (B)

48) 1 mol of N2O4 contains 2 mole of N2

Answer option (B) : 2

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