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
The chemical reaction would most likely be written as follows:
A + B = AB
We cannot simply use the usual method of converting grams to moles since we do not have any idea on what are the identities of A and B. The only method we could use is to use the law of conservation of mass where mass inflow in a process should be equal to the mass out in the process. The total inflow of mass would be the mass of A and B and the outflow would be the product AB.
mass of A + mass of B = mass of AB
10.0 g A + 10.0 g B = mass of AB
mass of AB = 20.0 g
Answer:
sodium/potassium/rubidium/caesium/francium
Explanation:
all are group I elements, so they all have similar properties
D.because the water is needed to wet the sand and keep the reaction safe
