OD IS THE ANSWER
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1) If we had excess Cl2, then limiting reagent is Al
so moles of Al will be the same as the moles of AlCl3 produced at the output
moles of Al = 23/27 = 0.8518 moles
2)If we had excess Al , then limiting reagent is Cl2
so moles of Cl2 will be the 3/2 times as the moles of AlCl3 produced at the output
moles of Cl2 = 28/71 = 0.394 moles
<span>hence moles of AlCl3 = 0.394*2/3 = 0.2629 moles</span>
<h3>
Answer:</h3>
Lead-205 (Pb-205)
<h3>
Explanation:</h3>
<u>We are given;</u>
We are supposed to identify its product after an alpha decay;
- Polonium-209 has a mass number of 209 and an atomic number of 84.
- When an element undergoes an alpha decay, the mass number decreases by 4 while the atomic number decreases by 2.
- Therefore, when Po-209 undergoes alpha decay it results to the formation of a product with a mass number of 205 and atomic number of 82.
- The product from this decay is Pb-205, because Pb-205 has a mass number of 205 and atomic number 82.
- The equation for the decay is;
²⁰⁹₈₄Po → ²⁰⁵₈₂Pb + ⁴₂He
- Note; An alpha particle is represented by a helium nucleus, ⁴₂He.
Answer:
6.5 mL
Explanation:
Step 1: Write the balanced reaction
Ca(OH)₂ + 2 HNO₃ ⇒ Ca(NO₃)₂ + 2 H₂O
Step 2: Calculate the reacting moles of nitric acid
25.0 mL of 0.50 M nitric acid react.
Step 3: Calculate the reacting moles of calcium hydroxide
The molar ratio of Ca(OH)₂ to HNO₃ is 1:2. The reacting moles of Ca(OH)₂ are 1/2 × 0.013 mol = 6.5 × 10⁻³ mol
Step 4: Calculate the volume of calcium hydroxide
To answer this, we need the concentration of calcium hydroxide. Since the data is missing, let's suppose it is 1.0 M.