The image below represents which of the following (Check all that apply):

Balance the following redox reaction occurring in an acidic solution. The coefficient of Cr2O72−(aq) is given. Enter the coeffic

ollegr [7]

Answer:

Cr₂O₇²⁻ (aq) + 6Ti³⁺ (aq) + 2 H⁺(aq) → 2 Cr³⁺ (aq) + 6TiO²⁺(aq) + H2O(l)

Explanation:

Cr₂O₇²⁻ (aq) + Ti³⁺ (aq) + H⁺ (aq) → Cr³⁺ (aq) + TiO²⁺ (aq) + H₂O(l)

This is the reaction, without stoichiometry.

We have to notice the oxidation number of each element.

In dicromate, Cr acts with +6 and we have Cr³⁺, so oxidation number has decreased. .- REDUCTION

Ti³⁺ acts with +3, and in TiO²⁺ acts with +4 so oxidation number has increased. .- OXIDATION

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O

To change 6+ to 3+, Cr had to lose 3 e⁻, but as we have two Cr, it has lost 6e⁻. As we have 7 Oxygens in reactant side, we have to add water, as the same amount of oxygens atoms we have, in products side. Finally, we have to add protons in reactant side, to ballance the H.

H₂O + Ti³⁺ → TiO²⁺ + 1e⁻ + 2H⁺

Titanium has to win 1 e⁻ to change 3+ to 4+. We had to add 1 water in reactant, and 2H⁺ in products, to get all the half reaction ballanced.

Now we have to ballance the electrons, so we can cancel them.

(14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O) .1

(H₂O + Ti³⁺ → TiO²⁺ + 1e⁻ + 2H⁺) .6

Wre multiply, second half reaction .6

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O

6H₂O + 6Ti³⁺ → 6TiO²⁺ + 6e⁻ + 12H⁺

Now we can sum, the half reactions:

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ + 6H₂O + 6Ti³⁺ → 6TiO²⁺ + 6e⁻ + 12H⁺ + 2Cr³⁺ + 7H₂O

Electrons are cancelled and we can also operate with water and protons

7H₂O - 6H₂O = H₂O

14 H⁺ - 12H⁺ = 2H⁺

The final ballanced equation is:

2H⁺ + Cr₂O₇²⁻ + 6Ti³⁺ → 6TiO²⁺ + 2Cr³⁺ + H₂O

8 0

2 years ago

(g) On the graph in part (d) , carefully draw a curve that shows the results of the second titration, in which the student titra

atroni [7]

Answer:

Here's what I get

Explanation:

(g) Titration curves

I can't draw two curves on the same graph, but I can draw two separate curves for you.

The graph in part (d) had an equivalence point at 20 mL.

In the second titration, the NaOH was twice as concentrated, so the volume to equivalence point would be half as much — 10 mL.

The two titration curves are below.

(h) Evidence of reaction

HCl and NaOH are both colourless.

They don't evolve a gas or form a precipitate when they react.

The student probably noticed that the Erlenmeyer flask warmed up — a sign of a chemical change.

4 0

2 years ago

What are the intermolecular forces between molecules in a liquid sample of sulfur trioxide,SO3?a. hydrogen bonding b. dipole-dip

fgiga [73]

Answer:

dispersion forces

Explanation:

SO3 is a trigonal planar molecule. All the dipoles of the S-O bonds cancel out making the molecule to be a nonpolar molecule.

The primary intermolecular force in nonpolar molecules is the London dispersion forces. As expected, the London dispersion forces is the intermolecular force present in SO3.

Hence SO3 is a symmetrical molecule having only weak dispersion forces acting between its molecules.

8 0

2 years ago

Copper(II) sulfide, CuS, is used in the development of aniline black dye in textile printing. What is the maximum mass of CuS wh

Naya [18.7K]

Answer:

1.82 g is the maximum mass of CuS.

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

Molarity = 0.500 M

Volume = 38.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 38.0×10⁻³ L

Thus, moles of $CuCl_2$ :

$Moles=0.500 \times {38.0\times 10^{-3}}\ moles$

<u>Moles of $CuCl_2$ = 0.019 moles </u>

Molarity = 0.600 M

Volume = 42.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 42.0×10⁻³ L

Thus, moles of $(NH_4)_2S$ :

$Moles=0.600 \times {42.0\times 10^{-3}}\ moles$

<u>Moles of $(NH_4)_2S$ = 0.0252 moles </u>

According to the given reaction:

$CuCl_2_{(aq)}+(NH_4)_2S_{(aq)}\rightarrow CuS_{(s)}+2NH_4Cl_{(aq)}$

1 mole of $CuCl_2$ reacts with 1 mole of $(NH_4)_2S$

So,

0.019 mole of $CuCl_2$ reacts with 0.019 mole of $(NH_4)_2S$

Moles of $(NH_4)_2S$ = 0.019 mole

Available moles of $(NH_4)_2S$ = 0.0252 mole

<u>Limiting reagent is the one which is present in small amount. Thus, $CuCl_2$ is limiting reagent.</u>

The formation of the product is governed by the limiting reagent. So,

1 mole of $CuCl_2$ gives 1 mole of $CuS$

0.019 mole of $CuCl_2$ gives 0.019 mole of $CuS$

Moles of $CuS$ formed = 0.019 moles

Molar mass of $CuS$ = 95.611 g/mol

Mass of $CuS$ = Moles × Molar mass = 0.019 × 95.611 g = 1.82 g

<u>1.82 g is the maximum mass of CuS.</u>

5 0

2 years ago

Which of the compounds above are strong enough acids to react almost completely with a hydroxide ion (pka of h2o = 15.74) or wit

luda_lava [24]

The compounds can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Further explanation </em></h3>

In an acid-base reaction, it can be determined whether or not a reaction occurs by knowing the value of pKa or Ka from acid and conjugate acid (acid from the reaction)

Acids and bases according to Bronsted-Lowry

Acid = donor (donor) proton (H + ion)

Base = proton (receiver) acceptor (H + ion)

If the acid gives (H +), then the remaining acid is a conjugate base because it accepts protons. Conversely, if a base receives (H +), then the base formed can release protons and is called the conjugate acid from the original base.

From this, it can be seen whether the acid in the product can give its proton to a base (or acid which has a lower Ka value) so that the reaction can go to the right to produce the product.

The step that needs to be done is to know the pKa value of the two acids (one on the left side and one on the right side of the arrow), then just determine the value of the equilibrium constant

Can be formulated:

K acid-base reaction = Ka acid on the left : K acid on the right.

or:

pK = acid pKa on the left - pKa acid on the right

K = equilibrium constant for acid-base reactions

pK = -log K;

$K~=~10^{-pK}$