> 2,000
mL of a 5.0 × 10–5% (w/v) sucrose solution
5.0 × 10–3
g/mL * 2000 mL * (1 mol / 342.30 g) = 0.0292 mol
<span>
> 2,000 mL of a 5.0 ppm sucrose solution</span>
5 grams /
1000000 mL * 2000 mL* (1 mol / 342.30 g) = 0.0000292 mol
<span>
> 20 mL of a 5.0 M sucrose solution </span>
5.0 M *
0.020 L = 0.1 mol
Answer:
<span>2,000 mL
of a 5.0 ppm sucrose solution</span>
Answer: B. It’s a dilute strong base.
Explanation:
1) Definition of acids and bases: as per Bronsted-Lowry model, an acid is a substance that donates hydrogen ions and a base is a substance that accepts hydrogen ions.
Ca(OH)₂ does not have hydrogen ions to donate, but it can accept hydrogen ions to form H₂O according to this equation: H⁺ + OH⁻ → H₂O.
Hence, Ca(OH)₂ is a base.
2) Definition of strong base: a strong base is a base that dissociates completely into metallic and hydroxide ions in aqueous solutions, while a weak base dissociates partially.
Hence, Ca(OH)₂ is a strong base.
3) Definition of dilute: it refers to a solution meaning that the substance is not pure and the concentration is low. Since, the solution the Ca(OH)₂ is 0.02 M means that it is dilute.
Therefore, we have found that the description of 0.02 M Ca(OH)₂ is that is is a dilute strong base (option B).
The formula for pH given the pKa and the concentrations
are:
pH = pKa + log [a–]/[ha]
<span>
Therefore calculating:</span>
3.75 = 3.75 + log [a–]/[ha]
log [a–]/[ha] = 0
[a–]/[ha] = 10^0
<span>[a–]/[ha] = 1</span>
5.2e+26is it this? im not sure
Answer:
can you provide more to the question
Explanation:
