Answer:
Ag₂CrO₄(s) + H⁺(aq) ⟶ 2Ag⁺(aq) + HCrO₄⁻(aq)
Explanation:
Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq).
Silver chromate is the salt of a strong base (AgOH) and a weak acid (H₂CrO₄).
HCrO₄⁻ is an even weaker acid than H₂CrO₄, so CrO₄²⁻ is a strong base.
Any added H⁺ will immediately combine with the chromate ions according to the reaction
H⁺ + CrO₄²⁻ ⟶ HCrO₄⁻
thereby removing chromate ions from solution.
According to Le Châtelier's Principle, more silver chromate will dissolve to replace the chromate ions that the H⁺ removes.
The overall equation for the reaction is
Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + <em>CrO₄²⁻(aq)
</em>
<u>H⁺(aq) + </u><em><u>CrO₄²⁻(aq)</u></em><u> ⟶ HCrO₄⁻(aq)
</u>
Ag₂CrO₄(s) + H⁺(aq) ⟶ 2Ag⁺(aq) + HCrO₄⁻(aq)
Answer:
Mitochondria are the "powerhouses" of the cell, breaking down fuel molecules and capturing energy in cellular respiration. Chloroplasts are found in plants and algae. They're responsible for capturing light energy to make sugars in photosynthesis.
Answer:
Explanation:
Hello!
In this case, since the sulfuric acid and sodium hydroxide react in a 1:2 mole ratio, given the reaction, we realize they have the following mole ratio at the equivalence point:
Which in terms of concentrations and volumes is:
Thus, we can plug in the volume and concentration of acid to find the moles of base:
Best regards!
Answer::
<u><em>Lower rate of reaction</em></u>
Explanation:
Lower concentration of reactant in an experiment is indicative of fewer ions or atoms present, which means a slower rate of reaction. In a titration reaction for instance, the end point would take longer to be substantiated, thereby increasing volume of titres. This is further backed up by collision theory which states that more particles in a system improves combinations of molecules.
This however may not be the case if the iodine in question is catalyst, in which case the change in concentration has no effect whatsoever on the reaction rte.
I hope this explanation fits the brief.
Answer:
The concentration of chloride ion is 2.887 M.
Explanation:
Here we have a combination BaCl₂ and AlCl₃
1 mole BaCl₂ can produce 2 moles chloride ions as follows;
BaCl₂ → Ba²⁺ + 2Cl⁻
Therefore;
0.554 mole BaCl₂ can produce 2×0.554 moles chloride ions;
1 mole AlCl₃ can produce 3 moles chloride ions as follows;
AlCl₃ → Al³⁺ + 3Cl⁻
Therefore;
0.593 mole AlCl₃ can produce 3×0.593 moles chloride ions
Total concentration of chloride ions = 2×0.554 + 3×0.593 = 2.887 moles of Cl⁻ in 2×16.8 ml solution
The concentration of Cl⁻ is 2.887 M
