Answer:
Q1) Distillation relies on evaporation to purify water
Explanation:
contaminated water is heated to form steam. inorganic compounds and large non-volatile organic molecules do not evaporate with the water and are left behind. The steam then cools and condenses to form purified water
Answer is: this is an example of an Arrhenius acid.
An Arrhenius acid is a
substance that dissociates in water to form hydrogen ions or protons (H⁺).
For example hydrochloric acid: HCl(aq) → H⁺(aq) + Cl⁻(aq).
An Arrhenius base is a
substance that dissociates in water to form hydroxide ions (OH⁻<span>).
In this example lithium hydroxide is an Arrhenius base:</span>
LiOH(aq) → Li⁺(aq) + OH⁻(aq).
I think it would be B maybe
Answer:
Explanation:
Combination reactions are the reactions in which two or more atoms/ molecules combine to give a single product.
Aluminum metal combines with iodine to given aluminium iodide.
Aluminium has a valency of 3 and iodine has valency of 1, so the formula is
The reaction is shown below as:
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
