Answer: 72.41% and 26.90% respectively.
Explanation:
At 60°C, you can dissolve 46.4g of acetanilide in 100mL of ethanol. If you lower the temperature, at 0°C, you can dissolve just 12.8g, which means (46.4g-12.8g)=33.6g of acetanilide must have precipitated from the solution.
We can calculate recovery as:
So the answer to the first question is 72.41%.
For the second part just use the same formula, the mass of the precipitate is the final mass minus the initial mass, (171mg-125mg)=46mg.
So the answer to the second question is 26.90%.
Answer:
[Ca⁺²] = 0.0123 M
1234 ppm of CaCO₃
Explanation:
Ca(In)²⁺ + EDTA ⟶ Ca(EDTA)²⁺ + In
To solve this problem we need to keep in mind the definition of Molarity (moles/L) and parts per million (mg CaCO₃ / L water). First we <u>calculate the moles of EDTA</u>:
- 0.0450 M * 13.70 mL = 0.6165 mmol EDTA
Looking at the reaction we see that one mol of EDTA reacts with one mol of Ca(In)²⁺, so we have as well 0.6165 mmol Ca(In)²⁺ (which for this case is the same as Ca⁺²). We calculate the molarity of Ca⁺²:
- [Ca⁺²] = 0.6165 mmol / 50 mL = 0.0123 M
For calculating the concentration in ppm, we use the moles of Ca⁺² and the molecular weight of CaCO₃ (100.09 g/mol):
- 0.6165 mmol Ca⁺² *
* 100.09 mg/mmol = 61.70 mg CaCO₃
<u>That is the mass of CaCO₃ present in 50 mL</u> (or 0.05 L) of water, so the concentration in ppm is:
- 61.70 mg / 0.05 L = 1234 ppm
The oxidation numbers of nitrogen in NH3, HNO3, and NO2 are, respectively: -3, -5, +4 +3, +5, +4 -3, +5, -4 -3, +5, +4
Evgesh-ka [11]
In NH3 , let oxidation number of N be x
x + (+1)3 = 0
x = -3
In HNO3 , let oxidation number of N be x
1 + x + (-2)3 = 0
x = +5
In NO2 , let oxidation number of N be x
x + (-2)2 = 0
x = +4
Tarnish is Ag2S-silver sulfide and the oxidation state of silver is +1
