Answer:
a. forces acting on the object
The empirical formula for a compound is KClO3
Explanation
find the moles of each element
moles = % composition/molar mass
molar mass of of potassium =39g/mol ,chlorine = 35.5 g/mol, oxygen =16 g/mol
moles of potassium = 31.9 / 39 = 0.818 moles
moles of chlorine = 28.9/35.5 = 0.814 moles
moles of oxygen = 39.2/ 16 = 2.45 moles
find the mole ratio by dividing with the smallest mole = 0.814 moles
potassium = 0.818/0.814 =1
chlorine = 0.814/0.814 = 1
oxygen = 2.45 /0.814 =3
the empirical formula is therefore = KClO3
Answer:
1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.
2) The pH of the solution after adding HCl is 12.6
Explanation:
10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.


There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.
NaOH + HCl ⇒ NaCl + H₂O
Initial 2.5 × 10⁻³ 1.5 × 10⁻³ 0 0
Reaction -1.5 × 10⁻³ -1.5 × 10⁻³ 1.5 × 10⁻³ 1.5 × 10⁻³
Final 1.0 × 10⁻³ 0 1.5 × 10⁻³ 1.5 × 10⁻³
The concentration of NaOH is:
![[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M](https://tex.z-dn.net/?f=%5BNaOH%5D%3D%5Cfrac%7B1.0%20%5Ctimes%2010%5E%7B-3%7D%20mol%20%7D%7B25.0%20%5Ctimes%2010%5E%7B-3%7D%20L%7D%20%3D0.040M)
NaOH is a strong base so [OH⁻] = [NaOH].
Finally, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log 0.040 = 1.4
pH = 14 - pOH = 14 - 1.4 = 12.6