A student was given the task of titrating a 20.mL sample of 0.10MHCl(aq) with 0.10MNaOH(aq) . The HCl(aq) was placed in an Erlenmeyer flask. An equation for the reaction that occurs during the titration is given above.

The box below to the left represents ions in a certain volume of 0.10MHCl(aq) . In the box below to the right, draw a representation of ions in the same volume of 0.20MHCl(aq)

Here's what I get  

Step-by-step Explanation

(a) Effect of dilution

There will be no effect on the volume of NaOH needed.

The amount of HCl will be halved, so the amount of NaOH will be halved.

However, the concentration of NaOH is also halved, so you will need twice the volume.

You will be back to the same volume as before dilution.

(b) Net ionic equation

Molecular: HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)

Ionic: H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + OH⁻(aq) ⟶ Na⁺(aq) + Cl⁻(aq) + H₂O(l)

Net ionic: H⁺(aq) + OH⁻(aq) ⟶ H₂O(l)

(c) Proton acceptor

H⁺ is the proton. OH⁻ accepts the proton and forms water.

(d) Moles of HCl

(e) Equivalence point

The equivalence point is the point at which the titration curve intersects the pH 7 line.

(f) Schematic representation

Assume the box for 0.10 mol·L⁻¹ HCl contains four black dots (H⁺) and four open circles (Cl⁻).

The 0.20 mol·L⁻¹ solution is twice as concentrated.

It will contain eight black dots and eight open circles.

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6

the atomic mass is the bottom number on the element.

-steel jelly