Answer:
Molarity: 0.522M
Percentage by mass: 2.36 (w/w) %
Explanation:
Formic acid, HCOOH reacts with NaOH as follows:
HCOOH + NaOH → NaCOOH + H₂O
To solve this question we must find the moles of NaOH added = Moles formic acid. Taken into account the dilution that was made we can find the moles -And molarity of formic acid and its percentage by mass as follows:
<em>Moles NaOH = Moles HCOOH:</em>
0.01580L * (0.1322mol / L) =0.002089 moles HCOOH
<em>Moles in the original solution:</em>
0.002089 moles HCOOH * (25mL / 10mL) = 0.005222 moles HCOOH
<em>Molarity of the solution:</em>
0.005222 moles HCOOH / 0.01000L =
<h3>0.522M</h3>
<em>Mass HCOOH in 1L -Molar mass: 46.03g/mol-</em>
0.522moles * (46.03g / mol) = 24.04g HCOOH
<em>Mass solution:</em>
1L = 1000mL * (1.02g / mL) = 1020g solution
<em>Mass percent:</em>
24.04g HCOOH / 1020g solution * 100
2.36 (w/w) %
Answer:
5.0x10⁻⁵ M
Explanation:
It seems the question is incomplete, however this is the data that has been found in a web search:
" One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample of silver nitrate solution. Any chloride anions in solution will combine with the silver cations to produce bright white silver chloride precipitate. Suppose a EPA chemist tests a 250 mL sample of groundwater known to be contaminated with nickel(II) chloride, which would react with silver nitrate solution like this:
NiCl₂ + 2AgNO₃ → 2AgCl + Ni(NO₃)₂
The chemist adds 50 mM silver nitrate solution to the sample until silver chloride stops forming. She then washes, dries, and weighs the precipitate. She finds she has collected 3.6 mg of silver chloride. Calculate the concentration of nickel(II) chloride contaminant in the original groundwater sample. Round your answer to 2 significant digits. "
Keep in mind that while the process is the same, if the values in your question are different, then your answer will be different as well.
First we <u>calculate the moles of nickel chloride found in the 250 mL sample</u>:
- 3.6 mg AgCl ÷ 143.32 mg/mmol * = 0.0126 mmol NiCl₂
Now we <u>divide the moles by the volume to calculate the molarity</u>:
- 0.0126 mmol / 250 mL = 5.0x10⁻⁵M
Answer:
U= 238g/mol
U2O5= 556g/mol
Explanation:
Since U= 238
O=16
U3O5= 2(238)+3(16)=556g/mol
Percentage recovery gives us an idea of the amount of pure substance recovered after the chemical reaction. Percentage recovery can be more than 100 % or less than 100 %. Usually, in any experiment performed the weight percentage recovery will be less than 100. Percent recovery values greater than 100 show that the recovered compound is contaminated.
Amount of acetaminophen initially taken = 350 mg
Amount of acetaminophen obtained after recovery =185 mg
=
= 52.9%
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