Answer:
4.4 g
Explanation:
Step 1: Write the balanced equation
Cu + 4 HNO₃ ⇒ Cu(NO₃)₂ + 2 NO₂ + 2 H₂O
Step 2: Calculate the moles corresponding to 3.2 L of NO₂ at STP
At standard temperature and pressure, 1 mole of NO₂ occupies 22.4 L.
3.2 L × 1 mol/22.4 L = 0.14 mol
Step 3: Calculate the moles of Cu needed to produce 0.14 moles of NO₂
The molar ratio of Cu to NO₂ is 1:2. The moles of Cu needed are 1/2 × 0.14 mol = 0.070 mol.
Step 4: Calculate the mass corresponding to 0.070 moles of Cu
The molar mass of Cu is 63.55 g/mol.
0.070 mol × 63.55 g/mol = 4.4 g
The answer is the first bubble, it absorbs white light and refracts green, this is because the color you see is the reflection of the object
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
Answer: it’s number 2 add a catalyst
Hopefully this helped :)
<span>Okay, a mole of potassium perchlorate contains 6.02x1023 formula units of potassium perchlorate, but you're asking about individual atoms. So, let's look at the formula: KClO3. That's 1 potassium, 1 chlorine, and 3 oxygens, for a total of 5 atoms per formula unit. Now, multiple 5 by Avogadro's number above, to get 30.1x1023, which simplifies to 3.01x1024 atoms.</span>
