Answer:
It is the third one
Explanation:
"Both atoms achieve a more stable configuration."
In the year 1909, Robert A. Millikan and Harvey Fletcher performed the oil drop experiment.
Answer:

Explanation:
Hello,
In this case, for this heat transfer process in which the heat lost by the hot platinum is gained by the cold deuterium oxide based on the equation:

We can represent the heats in terms of mass, heat capacities and temperatures:

Thus, we solve for the mass of platinum:

Next, by using the density of platinum we compute the volume:

Which computed in terms of the edge length is:

Therefore, the edge length turns out:
![a=\sqrt[3]{180cm^3}\\ \\a=5.65cm](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7B180cm%5E3%7D%5C%5C%20%5C%5Ca%3D5.65cm)
Best regards.
Answer:
ScCl₂
General Formulas and Concepts:
<u>Chemistry</u>
- Reading a Periodic Table
- Reaction Prediction
- Determining Chemical Compounds
Explanation:
<u>Step 1: Define</u>
Scandium (II)
Cl
<u>Step 2: Determine Charges</u>
Sc²⁺
Cl⁻
<u>Step 3: Predict Compound</u>
<em>We need to balance out the charges so the overall charge is 0.</em>
ScCl₂
<u>Step 4: Reaction</u>
RxN: Sc²⁺ + Cl₂ → ScCl₂
Answer:
Number of moles = 3.73 mol
Explanation:
Given data:
Mass of LiCl = 158 g
Number of moles = ?
Solution:
Formula:
Number of moles = mass/ molar mass
Molar mass of LiCl = 42.4 g/mol
Number of moles = 158 g / 42.4 g/mol
Number of moles = 3.73 mol