When non-metal atoms ionize they gain electrons.<span> Some examples of this are the halogen elements: F, Cl, Br and I, each, can gain one electron from the respective anions, F-, Cl-, Br-, and I-. O and S, may gain two electrons to form the anions O2- and S2-. This is due to the fact that those elements only need one (in the case of the halogens) or two (in the case of O and S) electrons to reach the most stable configuration of the closest noble gas (with the last shell of electrons full), so they are ready to gain those electrons and form the corresponding ions.</span>
Answer: sorry can't help ya
Explanation:
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Answer:
14 mol O₂
Explanation:
The reaction between CO and O₂ is the following:
CO + O₂ → CO₂
We balance the equation with a coefficient 2 in CO and CO₂ to obtain the same number of O atoms:
2CO + O₂ → 2CO₂
As we can see from the balanced equation, 1 mol of O₂ is required to react with 2 moles of CO. Thus, the conversion factor is 1 mol of O₂/2 mol CO. We multiply the moles of CO by the conversion factor to calculate the moles of O₂ that are required:
28 mol CO x 1 mol of O₂/2 mol CO = 14 mol O₂
Answer:
Assuming that all of the oxygen is used up, 1.53×4111.53×411 or 0.556 moles of C2H3Br3 are required. Because there are only 0.286 moles of C2H3Br3 available, C2H3Br3 is the limiting reagent.
Limiting Reagent What is the limiting reagent if 76.4 grams of C2H3Br3 were reacted with 49.1 grams of O2? C2H3Br3 + 11O2 → 8CO2 + 6H2O + 6Br2 SOLUTION Using Approach 1: A. 76.4g × (1 mol/ 266.72 g) = 0.286 moles C2H3Br3 49.1g × (1 mole/ 32 g) = 1.53 moles O2 B.
Explanation:
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https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Map%3A_Introductory_Chemistry_(Tro)/08%3A_Quantities_in_Chemical_Reactions/8.04%3A_Limiting_Reactant_and_Theoretical_Yield
