Where is the following statements??
HCl is a polar molecule with the hydrogen part being partial positive while the chlorine end being partial negative. This is because hydrogen has an electronegativity of 2.1, and chlorine has an electronegativity of 3.0. This means that chlorine attracted most of the electron cloud of molecule hence is the negative dipole, The dipole moment of HCl is 1.08 D (debyes). A Debye is equal to 3.34 x 10-30 coulomb-meters (C-m). The charge of each molecule is o.176+ for H and 0.176- for the Cl
Percentage recovery gives us an idea of the amount of pure substance recovered after the chemical reaction. Percentage recovery can be more than 100 % or less than 100 %. Usually, in any experiment performed the weight percentage recovery will be less than 100. Percent recovery values greater than 100 show that the recovered compound is contaminated.
Amount of acetaminophen initially taken = 350 mg
Amount of acetaminophen obtained after recovery =185 mg
= 
= 52.9%
Answer:
0.645 L
Explanation:
To find the volume, you need to (1) convert grams to moles (using the molar mass) and then (2) calculate the volume (using the molarity ratio). The final answer should have 3 sig figs to match the sig figs of the given values.
(Step 1)
Molar Mass (KOH): 39.098 g/mol + 15.998 g/mol + 1.008 g/mol
Molar Mass (KOH): 56.104 g/mol
19.9 grams KOH 1 mole
-------------------------- x ----------------------- = 0.355 moles KOH
56.014 grams
(Step 2)
Molarity = moles / volume <----- Molarity ratio
0.550 M = 0.355 moles / volume <----- Insert values
(0.550 M) x volume = 0.355 moles <----- Multiply both sides by volume
volume = 0.645 L <----- Divide both sides by 0.550
The volume of the 0.279 M Ca(OH)₂ solution required to neutralize 24.5 mL of 0.390 M H₃PO₄ is 51.4 mL
<h3>Balanced equation </h3>
2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O
From the balanced equation above,
- The mole ratio of the acid, H₃PO₄ (nA) = 2
- The mole ratio of the base, Ca(OH)₂ (nB) = 3
<h3>How to determine the volume of Ca(OH)₂ </h3>
- Molarity of acid, H₃PO₄ (Ma) = 0.390 M
- Volume of acid, H₃PO₄ (Va) = 24.5 mL
- Molarity of base, Ca(OH)₂ (Mb) = 0.279 M
- Volume of base, Ca(OH)₂ (Vb) =?
MaVa / MbVb = nA / nB
(0.39 × 24.5) / (0.279 × Vb) = 2/3
9.555 / (0.279 × Vb) = 2/3
Cross multiply
2 × 0.279 × Vb = 9.555 × 3
0.558 × Vb = 28.665
Divide both side by 0.558
Vb = 28.665 / 0.558
Vb = 51.4 mL
Thus, the volume of the Ca(OH)₂ solution needed is 51.4 mL
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