Answer:
The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.
Explanation:
Using Boyle's law

Given ,
V₁ = 3.6 L
V₂ = ?
P₁ = 1.0 atm
P₂ = 13.3 atm (From correct source)
Using above equation as:




The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.
Answer: Heating a crucible to remove water from a hydrate.
Explanation:
The options are:
a. Heating a solvent to help a solute dissolve.
b. Heating an isolated solid to dry it.
c. Heating water to boiling for a water bath.
d. Heating a crucible to remove water from a hydrate.
The procedure that can be performed on a hot plate are:
a. Heating a solvent to help a solute dissolve.
b. Heating an isolated solid to dry it.
c. Heating water to boiling for a water bath.
It should be noted that the hot plate cannot be used for heating of crucible in order to remove water from a hydrate. It is not advisable for someone to heat any silica or ceramic objects on a hot plate.
Therefore, heating a crucible to remove water from a hydrate is the correct option.
The balanced equation for the above reaction is
HBr + KOH ---> KBr + H₂O
stoichiometry of HBr to KOH is 1:1
HBr is a strong acid and KOH is a strong base and they both completely dissociate.
The number of HBr moles present - 0.25 M / 1000 mL/L x 52.0 mL = 0.013 mol
The number of KOH moles added - 0.50 M / 1000 mL/L x 26.0 mL = 0.013 mol
the number of H⁺ ions = number of OH⁻ ions
therefore complete neutralisation occurs.
Therefore solution is neutral. At 25 °C, when the solution is neutral, pH = 7.
Then pH of solution is 7