Explanation:
Can you please make the question more precise..
Thanks..
Answer:
In comparison to Part 1 of this experiment, we observed similar reactions when determining the make up of our unknown. When testing for Mn2+ we observed a color change that resulted in a darker brown/red color, when testing for Co2+ we observed the formation of foamy bubbles but we could not conclude that a gas had formed, when testing for Fe3+ the result was a liquid red in color, when testing for Cr3+ we observed no change, when testing for Zn2+ we observed the formation of a pink/red liquid, when testing for K+ we observed the formation of a precipitate, when testing for Ca2+ we observe the formation of a precipitate. Sources of error may have occurred when observing whether or not an actual reaction had taken place or not, using glassware that wasn't fully cleaned, or the accidental mix of various other liquids in the lab
Explanation:
Answer:
The mass percent of Al(OH)₃ is 15.3%
Explanation:
The reaction is:
Al(OH)₃ + 3HCl = AlCl₃ + 3H₂O
The excess acid is neutralized with a solution of sodium hidroxide, in the reaction:
NaOH + HCl = NaCl + H₂O
The total moles of HCl is:
From the second titration, the moles of excess of HCl is:
The difference between the total and excess of HCl, it can be know the moles that reacts with the aluminum hydroxide, is:
The ratio between HCl and Al(OH)₃ is 3:1. The MW for aluminum hydroxide is 78 g/mol, thus:
The percentage of Al(OH)₃ is:
%
Answer:
Cl2, because it gained electrons
Explanation:
2Fe^2+ went from 2+ to 3+
Cl2 went from 0 to -1
Reduction-gain
Oxidation-loss
- means gained and + means loss trust me it doesn't make sense but if you stick to it it'll be a lot easier for you.
Answer:
0.210 M
Explanation:
<em>A 75.0 mL aliquot of a 1.70 M solution is diluted to a total volume of 278 mL.</em>
In order to find out the resulting concentration (C₂) we will use the dilution rule.
C₁ × V₁ = C₂ × V₂
1.70 M × 75.0 mL = C₂ × 278 mL
C₂ = 0.459 M
<em>A 139 mL portion of that solution is diluted by adding 165 mL of water. What is the final concentration? Assume the volumes are additive.</em>
Since the volumes are additive, the final volume V₂ is 139 mL + 165 mL = 304 mL. Next, we can use the dilution rule.
C₁ × V₁ = C₂ × V₂
0.459 M × 139 mL = C₂ × 304 mL
C₂ = 0.210 M
