A salt solution is added to a marine aquarium. What mass of sodium chloride is n

Question

A salt solution is added to a marine aquarium. What mass of sodium chloride is needed to produce 250.0 mL of a SOLUTION that has a concentration of 0.0500 M? If 50.0 mL of a 0.10 A/solution of sodium chloride is mixed with 50.0 mL of 0.10 Mmagnesium chloride, what is the molar concentration of chloride in the resulting solution? A 500 mg dietary supplement of L-lysine (an amino acid, 146.2 g/mol) required 68.4 mL of 0.100 A/NaOH to reach the end point. How many protons were removed for each L-lysine molecule in this titration? If one regular antacid tablet contains 500 mg of solid CaC03 (100 g/mol), how many mL of 1.0 M stomach acid (HC1) could it neutralize? Which of the following ionic compounds is insoluble in water? When the oxidation-reduction reaction shown here is balanced, how many electrons are transferred for each atom of copper that reacts? Methane (CH4) is a suitable fuel for burning because it is readily oxidized by oxygen gas, forming carbon dioxide and water. Hydrogen is also a good fuel for burning with oxygen gas, forming water as a product. What change in oxidation number always accompanies oxidation?

Explanation / Answer

1. a. 15M

Density =0.90 g/mL

Volume =1000 ml (since molarity is the number of moles in 1000ml of the solution)

Therefore, mass of the entire solution = density * volume = 900 gm

However mass % of ammonia in the solution = 28% =0.28 * 900 = 252gm

This implies 1000ml of the solution contains 252gm of ammonia

Molecular mass of ammonia =17gm

Therefore, Molarity = 252/17 = 14.82 M = 15M (appx)

17. b . 731 mg

Molarity =0.05M (0.05 moles in 1000ml of solution)

Volume of solution = 250 ml (0.05/4 moles in 250 ml of solution)

Molecular mass of NaCl = 58.5gm

Mass of Nacl to be added = m = (0.05 * 58.5)/4 =731mg

18. d. 0.15M

50 ml of 0.1M NaCl solution contains (0.1/1000)*50 = 0.005 moles of chloride ions

50ml of 0.1M MgCl2 solution contains (0.1/1000)*50*2 = 0.01 moles of chloride ions (MgCl2   gives two chloride ions)

Therefore 100ml of the resultant solution contains 0.015 moles of chloride ions

Molarity = (0.015/100)*1000 = 0.15M

19. b. 2

No. of moles of lysine = (500 * 10-3) /146.2 = 3.42 x 10-3

No. of moles of NaOH consumed = (68.4 * 0.1) / 1000 = 6.84 x 10-3

Each mole of NaOH gives 1 mole of OH- ions which requires 1mole of protons

Therefore no. of protons given by each lysine molecule = (6.84 x 10-3) / (3.42 x 10-3 ) = 2

20.b. 10ml

No. of moles of CaCO3 in a single tablet = (500 * 10-3)/100 = 0.005 moles

Each mole of CaCO3 consumes 2 moles of HCl to neutralize and forms H2CO3 and CaCl2

Therefore, 0.005 moles of CaCO3 will neutralize 0.005 * 2 = 0.01 moles of HCl

1.0M HCl solution contains 1mole of HCl in 1000 ml

10ml of 1.0M HCl solution will contain (10/1000) =0.01 moles of HCl

21.b. BaSO4

22.b.2

The oxidation and reduction half reactions are

i. Ag+ + 1e- --------------------------- Ag

ii. Cu -------------------------------- Cu2+ + 2 e-

To balance the two equations we multiply i. with 2

2Ag+ + 2e- --------------------------- 2 Ag

Thus for each Cu atom 2 electrons are transferred

23.b. an increase

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