Response:
0.8853 mL
Clarification:
Initially, we convert 13 lb to kg, remembering that 1 lb = 0.454 kg:
- 13 lb *
= 5.902 kg
Next, we determine the required mg of acetaminophen to administer, applying the recommended dosage and infant's weight:
- 15 mg/kg * 5.902 kg = 88.53 mg
Finally, we compute the necessary mL of suspension, utilizing its concentration:
- 88.53 mg ÷ (80 mg/0.80 mL) = 0.8853 mL
The Chemistry Regents is one of the four science Regents exams. The remaining three are Earth Science, Living Environment, and Physics. Passing at least one of these four exams is a requirement for high school graduation.
(c) Cu + S → CuS is classified as a redox reaction
Explanation:
The following reactions are presented:
(a) K₂CrO₄ + BaCl₂ → BaCrO₄ + 2 KCl
(b) Pb²⁺ + 2 Br⁻ → PbBr₂
(c) Cu + S → CuS
Reaction (c) represents a redox reaction, as the oxidation states of the elements are changing. In this case:
Cu + S → CuS
In its elemental form, Cu has an oxidation state of 0, while in CuS (copper sulfide), its oxidation state changes to +2.
Similarly, S in its elemental form has an oxidation state of 0 and is -2 in CuS (copper sulfide).
Learn more about:
redox reactions
Answer: The correct selection is (b).
Explanation:
The energy required to detach an electron from an atom or ion in its gaseous state is termed ionization energy.
This indicates that a smaller atom necessitates a greater amount of energy to remove its valence electron. The reason for this is that there exists a strong attraction between the nucleus and the electrons in smaller atoms or elements.
Therefore, a significant amount of energy is needed to dislodge the valence electrons.
The electronic configuration for helium is 
. Hence, due to its fully occupied valence shell, it exhibits greater stability.
Consequently, a large amount of energy is needed to remove an electron from a helium atom.
In conclusion, from the choices provided, the ionization energy of helium will be greater than that of the diatomic molecule.
Convert 55.0g Ca(OH)2 to moles.
The calculation shows that 55.0g of Ca(OH)2 corresponds to 0.742 moles.
To find the volume, divide 0.742 mol of Ca(OH)2 by its molarity of 0.680M, yielding approximately 1.09L of Ca(OH)2.
If you disregard the negligible volume of the Ca(OH)2 itself, the resulting total volume of a 0.680M solution created by dissolving 55.0g of Ca(OH)2 in an appropriate amount of water would be 1.09L.
