Please provide the exact problem from Gina Wilson All Things Algebra Unit 2 Homework 5 so I can give you the correct answer.
This question needs more information to provide a complete answer. Missing Details Specify: The list of sentences to pair. Clarify: What is the pairing criterion (meaning, synonyms/antonyms, tense, logical connection)? Provide: The actual sentences or the worksheet data. Try Asking “Please provide the sentences to pair and the pairing rule (e.g., match meaning, opposite meaning, […]
The carbonate ion, CO₃²⁻, has a central carbon bonded to three oxygens in a trigonal planar arrangement; it has three resonance structures (one C=O and two C–O⁻ bonds) giving an overall -2 charge.
Central carbon with three oxygens in a trigonal plane; in one canonical Lewis form, one O is double-bonded to C and the other two are single-bonded with a -1 charge each. The carbonate ion is a resonance hybrid of three equivalent forms, giving each C–O bond a bond order of about 1.33.
Hydrogen carbonate (bicarbonate) is HCO₃⁻. Its Lewis structure has carbon at the center with three oxygens: one OH group, one double-bonded O, and one singly bonded O⁻; the negative charge is delocalized between the two non‑OH oxygens via resonance.
CO₃^{2-} carbonate ion: carbon in the center with three oxygens in a trigonal planar arrangement; three resonance structures (one C=O and two C–O bonds each) give an average C–O bond order of 4/3 and an overall -2 charge.
$$CO_{3}^{2-}$$ (carbonate) has three equivalent Lewis resonance structures with one C=O and two C–O^{-} bonds, and the negative charge is delocalized over the three oxygens.
HCO₃⁻ (bicarbonate) Lewis structure: HO–C(=O)–O⁻; the negative charge is delocalized over the oxygens by resonance.
Hydrogen carbonate (bicarbonate) has the Lewis structure HO–C(=O)–O^−, with the negative charge delocalized over the two non-hydroxyl oxygens by resonance.
CO3^{2-} (carbonate) has a central C bonded to three O atoms: one C=O bond and two C–O bonds; the two singly bonded oxygens carry -1 each, giving -2 total, with three resonance structures placing the double bond on different O atoms.
Continuously compounded interest accumulates at every instant, with amount given by $$A = P e^{rt}$$, where P is principal, r is the annual rate, and t is time in years.
Continuous compounding means interest is added continuously, not at discrete intervals. The amount is $$A = P e^{rt}$$, where P is the principal, r the rate, and t the time.
Continuous compounding means interest is added an infinite number of times per period, so the amount is A = P e^{rt}.
Continuous interest, or continuous compounding, is interest that accrues continuously, with amount A = P e^{rt} , where P is the principal, r the annual rate, and t the time in years.
Continuous compounding is interest added continuously, so the amount is A = P e^{rt}.
A = P e^{δ t}, where δ is the force of interest. Key Concepts Concept: Force of interest Concept: Continuous compounding Concept: Effective vs. nominal rates Steps Action: Identify P, δ, t Action: Compute A = P e^{δ t} Action: Compute I = A − P and i_eff = e^{δ} − 1 Formula $$A = […]
Interest compounded continuously means the amount grows as A = P e^{rt}, where P is the principal, r is the annual rate, and t is time.
A = P e^{rt} is the continuous compounding formula. Key Concepts Concept: Principal amount P Concept: Rate r as decimal Concept: Time t in years Steps Action: Identify P, r, t Action: Compute rt and evaluate e^{rt} Action: Compute A = P e^{rt} Formula $$A = P e^{rt}$$ Example Example: If P = 1000, r […]
The continuous compounding formula is A = P e^{rt}. Key Concepts Concept: Continuous compounding Concept: Principal, rate, time Concept: Exponential growth (Euler’s e) Steps Action: Identify P, r, t Action: Compute A = P e^{rt} Action: Interpret result: A ≥ P if r,t ≥ 0 Formula $$A = P e^{rt} quad text{where } A text{ […]