Question

Q3. When sodium sulfate and barium nitrate react, barium sulfate and sodium nitrate are formed. What mass of barium sulfate is obtained when 100.0 mL of 0.105 M sodium sulfate is added to 50.0 mL of 0.0850 M barium nitrate?   
A) 0.992 g    

B) 2.45 g    

C) 2.98 g    

D) 3.68 g    

E) 1.98 g

Q4. Write the net ionic equation for the reaction between hypochlorous acid and sodium hydroxide?

Q6. Suppose you have 100.00 ml of a solution of a dye and transfer 2.00 ml of the solution to a 100.00ml volumetric flask. After adding water to the 100.00 ml mark, you take 5.00 ml of that solution and again dilute to 100.00 ml. If you find the dye concentration in the final diluted sample is 0.000158 M, what was the dye concentration in the original solution.  

Q8. Select the net ionic equation for the reaction between sodium chloride and lead (II) nitrate.

A) NaCl(s) → Na⁺(aq) + Cl⁻(aq)

B) Na⁺(aq) + NO₃⁻(aq) → NaNO₃(s)

C) Pb(NO₃)₂(s) → Pb²⁺(aq) + 2NO₃⁻(aq)

D) Pb²⁺(aq) + 2Cl⁻(aq) → PbCl₂(s)

E) 2 NaCl(aq) + Pb(NO₃)₂(aq) → 2NaNO₃(s) + PbCl₂(s)

Answer 1

Q.4) Solution:

HClO is a weak acid and does not produce much H+ or ClO- ions, so it is shown unionized in an ionic reaction.

Complete ionic equation:

HClO(aq) + Na+(aq) + OH-(aq) ------> Na+(aq) + ClO-(aq) + H2O(l)

Net ionic equation (eliminate ions that appear on both sides equally:

HClO(aq) + OH-(aq) ------> ClO-(aq) + H2O(l) .

Q8).

Pb(NO₃)₂(aq) + 2 NaCl(aq) PbCl₂(s) + 2 NaNO₃(aq)

This complete equation may be rewritten in ionic form by using the solubility rules. lead(II) nitrate is soluble and therefore dissociated. the same about NaCl. As products, sodium nitrate is predicted to be soluble and will be dissociated. The lead(II) chloride, however, is insoluble . The above equation written in dissociated form is:

Pb²⁺(aq) + 2 NO₃^-(aq) + 2 Na⁺(aq) + 2 Cl^-(aq)    PbCl₂(s) + 2 Na⁺(aq) + 2 NO₃^-(aq)

At this point, one may cancel out those ions which have not participated in the reaction. Notice how the nitrate ions and sodium ions remain unchanged on both sides of the reaction.

Pb²⁺(aq) +  2 NO₃^-(aq) +  2 Na⁺(aq) + 2 Cl^-(aq)    PbCl₂(s) +  2 Na⁺(aq) +  2 NO₃^-(aq)

What remains is the net ionic equation, showing only those chemical species participating in a chemical process:

Pb²⁺(aq) + 2 Cl^-(aq) PbCl₂(s)

Option D - Correct Answer.

ANSWER Q3 AND Q4 are attach with photo. (IF u like the answer please give rate)