<h2><em><u>Technically yes but no because you have to fill the 3s orbital before the 5s orbital. </u></em><em><u>✨</u></em><em><u>✨</u></em><em><u>✨</u></em><em><u>✨</u></em><em><u>✨</u></em><em><u>✨</u></em><em><u>✨</u></em><em><u>✨</u></em><em><u>✨</u></em><em><u>✨</u></em><em><u>✨</u></em></h2>
Answer: The answer is 1/4
Explanation:
Answer:
I'm pretty sure this doesn't really connect with chemistry much but the answer should be <u>Schedule 1 </u> as stated by the Drug enforcement agency.
Answer:
All refineries have three basic steps: separation, conversion and treatment. During the separation process, the liquids and vapors separate into petroleum components called factions based on their weight and boiling point in distillation units.
Explanation:
Answer:
7.16 × 10⁻³ M
Explanation:
Let's consider the reduction reaction of copper during the electroplating.
Cu²⁺(aq) + 2 e⁻ ⇒ Cu(s)
We can calculate the moles of Cu²⁺ present in the solution using the following relations.
- 1 mole of electrons has a charge of 96486 C (Faraday's constant).
- 1 mole of Cu²⁺ is reduced when 2 moles of electrons are gained.
The moles of Cu²⁺ reduced are:
![2.30 min \times \frac{60s}{1min} \times \frac{3.00C}{s} \times \frac{1mole^{-} }{96486C} \times \frac{1molCu^{2+} }{2mole^{-} } = 2.15 \times 10^{-3} molCu^{2+}](https://tex.z-dn.net/?f=2.30%20min%20%5Ctimes%20%5Cfrac%7B60s%7D%7B1min%7D%20%5Ctimes%20%5Cfrac%7B3.00C%7D%7Bs%7D%20%5Ctimes%20%5Cfrac%7B1mole%5E%7B-%7D%20%7D%7B96486C%7D%20%5Ctimes%20%5Cfrac%7B1molCu%5E%7B2%2B%7D%20%7D%7B2mole%5E%7B-%7D%20%7D%20%3D%202.15%20%5Ctimes%2010%5E%7B-3%7D%20molCu%5E%7B2%2B%7D)
of Cu²⁺ are in 0.300 L of solution.
[Cu²⁺] = 2.15 × 10⁻³ mol/0.300 L = 7.16 × 10⁻³ M