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salantis [7]
3 years ago
10

Compare the solubility of silver chromate in each of the following aqueous solutions: Clear All 0.10 M AgCH3COO 0.10 M Na2CrO4 0

.10 M KCH3COO 0.10 M NH4NO3 More soluble than in pure water. Similar solubility as in pure water. Less soluble than in pure water.
Chemistry
1 answer:
slavikrds [6]3 years ago
4 0

Solution :

Comparing the solubility of silver chromate for the solutions :

$0.10 \ M \ AgCH_3COO$    -----     Less soluble than in pure water.

$0.10 \ M \ Na_2CrO_4$   ----- Less soluble than in pure water.

$0.10 \ M \ NH_4NO_3$   -----   Similar solubility as in the pure water

$0.10 \ M \ KCH_3COO$   -----   Similar solubility as in the pure water

The silver chromate dissociates to form :

$AgCrO_4 (s) \rightleftharpoons 2Ag^+ (aq) +CrO_4^{2-}(aq)$

When 0.1 M of $AgCH_3COO^-$ is added, the equilibrium shifts towards the reverse direction due to the common ion effect of Ag^+, so the solubility of Ag_2CrO_4 decreases.

Both AgCH_3COO and $KCH_3COO$ are neutral mediums, so they do not affect the solubility.

 

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The plant would be insect-resistant and improve its chances of success.

Explanation:

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Choose the ingredients needed for nuclear fusion. Check all that apply. energy helium gas high temperatures hydrogen gas low pre
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A chemist wants to extract copper metal from copper chloride solution. The chemist places 0.50 grams of aluminum foil in a solut
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Answer:

Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.

Explanation:

  • It is a stichiometry problem.
  • We should write the balance equation of the mentioned chemical reaction:

<em>2Al + 3CuCl₂ → 3Cu + 2AlCl₃.</em>

  • It is clear that 2.0 moles of Al foil reacts with 3.0 moles of CuCl₂ to produce 3.0 moles of Cu metal and 2.0 moles of AlCl₃.
  • Also, we need to calculate the number of moles of the reported masses of Al foil (0.50 g) and CuCl₂ (0.75 g) using the relation:

<em>n = mass / molar mass</em>

  • The no. of moles of Al foil = mass / atomic mass = (0.50 g) / (26.98 g/mol) = 0.0185 mol.
  • The no. of moles of CuCl₂ = mass / molar mass = (0.75 g) / (134.45 g/mol) = 5.578 x 10⁻³  mol.
  • <em>From the stichiometry Al foil reacts with CuCl₂ with a ratio of 2:3.</em>

∴ 3.85 x 10⁻³  mol of Al foil reacts completely with 5.578 x 10⁻³  mol of CuCl₂ with <em>(2:3)</em> ratio and CuCl₂ is the limiting reactant while Al foil is in excess.

  • From the stichiometry 3.0 moles of  CuCl₂ will produce the same no. of moles of copper metal (3.0 moles).
  • So, this reaction will produce 5.578 x 10⁻³ mol of copper metal.
  • Finally, we can calculate the mass of copper produced using:

mass of Cu = no. of moles x Atomic mass of Cu = (5.578 x 10⁻³  mol)(63.546 g/mol) = 0.354459 g ≅ 0.36 g.

  • <u><em>So, the answer is:</em></u>

<em>Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.</em>

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4 years ago
A vector is an arrow that shows both ___ and direction Of an object in motion
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the vector is an arrow that shows both <em><u>speed</u></em> and direction of an object

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Answer:

Here's what I get.

Explanation:

The MO diagrams of KrBr, XeCl, and XeBr are shown below.

They are similar, except for the numbering of the valence shell orbitals.

Also, I have drawn the s and p orbitals at the same energy levels for both atoms in the compounds. That is obviously not the case.

However, the MO diagrams are approximately correct.

The ground state electron configuration of KrF is

(1\sigma_{g})^{2}\, (1\sigma_{u}^{*})^{2} \, (2\sigma_{g})^{2} \, (2\sigma_{u}^{*})^{2} \, (3\sigma_{g})^{2} \,  (1\pi_{u})^{4} \, (1\pi_{g}^{*})^{4} \, (3\sigma_{g}^{*})^{1}

KrF⁺ will have one less electron than KrF.

You remove the antibonding electron from the highest energy orbital, so the bond order increases.

The KrF bond will be stronger.

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