Answer:
2.893 x 10⁻³ mol NaOH
[HCOOH] = 0.5786 mol/L
Explanation:
The balanced reaction equation is:
HCOOH + NaOH ⇒ NaHCOO + H₂O
At the endpoint in the titration, the amount of base added is just enough to react with all the formic acid present. So first we will calculate the moles of base added and use the molar ratio from the reaction equation to find the moles of formic acid that must have been present. Then we can find the concentration of formic acid.
The moles of base added is calculated as follows:
n = CV = (0.1088 mol/L)(26.59 mL) = 2.892992 mmol NaOH
Extra significant figures are kept to avoid round-off errors.
Now we relate the amount of NaOH to the amount of HCOOH through the molar ratio of 1:1.
(2.892992 mmol NaOH)(1 HCOOH/1 NaOH) = 2.892992 mmol HCOOH
The concentration of HCOOH to the correct number of significant figures is then calculated as follows:
C = n/V = (2.892992 mmol) / (5.00 mL) = 0.5786 mol/L
The question also asks to calculate the moles of base, so we convert millimoles to moles:
(2.892992 mmol NaOH)(1 mol/1000 mmol) = 2.893 x 10⁻³ mol NaOH
Answer:inform the lab instructor and get instructions
Explanation:
If you come across a chemical in the laboratory which has been wrongly labelled, do not be quick to dilute it or take any further action. The laboratory instructor who may have prepared the reagent himself or has better knowledge about the reagent should be contacted immediately so that he/she can give you instructions about what to do with the wrongly labelled reagent.
On addition to water<span> the Na+ section of </span>NaCl<span> is attracted to the oxygen side of the </span>water<span> molecules, while the Cl- side is attracted to the hydrogens' side of the </span>water<span> molecule. This causes the sodium chloride to split in </span>water<span>, and the </span>NaCl dissolves<span> into separate Na+ and Cl- atoms.</span>
1) Reactions which<span> involves both a </span>reduction<span> process and a complementary </span>oxidation<span> process are called </span>oxidation-reduction.they <span>can't be double replacement reactions.</span>
Hence,
WO3 + 3H2 ---> W + 3H2O is an oxidation-reduction reaction
The theoretical molar yield of lead (II) chloride will be 9 moles.
<h3>Stoichiometric calculation</h3>
First, we need to look at the equation of the reaction:
From the equation, the 1 mole of Pb2+ ion requires 2 moles of Cl- ion in order to produce 1 mole of lead (II) chloride.
Thus, with 18 moles Cl- ions, 9 moles of Pb2+ would be needed, instead of 12 moles. Pb2+ is simply in excess while Cl- can be said to be limiting.
Therefore, Cl- will determine how much of lead (II) chloride that will be produced. The ratio is 2 moles of Cl- to 1 mole of lead (II) chloride.
With 18 moles Cl-, 9 moles of lead (II) chloride will, thus, be produced.
More on mole ratios can be found here: brainly.com/question/14425689
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