Answer:
Scientists believe that convection circulation within the mantle helps continents to move. As heat from Earth's innermost layer—the core—transfers to the bottom layer of mantle rock, the rock warms, softens, and rises upward. ... This churning motion in the mantle appears to be a major factor in plate movement.
Explanation:
Explanation:
A non-electrolyte is defined as a solution which does not contain any ions and hence, it is unable to conduct electricity.
For example, when a non-polar substance like 
is dissolved in water then it will not dissociate into ions.
As electricity is the flow of ions or electrons. So, a non-electrolyte solution is not able to conduct electricity.
Similarly, a compound that is insoluble in water will not dissociate into ions. Hence, this type of solution will not be able to conduct electricity.
The correct answer is a metal atom forms a cation, and a nonmetal atom forms an anion. This is because metals are less electronegative than nonmetals and will therefore give electrons to nonmetals. Atoms that give up electrons will have a positive charge therefore becoming a cation while atoms that accept electrons will have a negative charge therefore becoming an anion.
Ions that have the same charge can't be attracted to each other since it takes a positive and negative charge to cause attractive forces.
A less electronegative atom will transfer electrons to a more electronegative atom.
A metal (cation) can pull electrons from another metal (not an ion) but that does not form an attractive force between the two metals (You will learn more about this when you go over reduction potentials, redox reactions, and electrochemistry).
I hope this helps. Let me know if anything is unclear.
Answer:
Here's what I get
Explanation:
Assume the initial concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.
We must calculate the initial concentration of HI.
1. We will need a chemical equation with concentrations, so let's gather all the information in one place.
H₂ + I₂ ⇌ 2HI
I/mol·L⁻¹: 0.30 0.15 x
2. Calculate the concentration of HI
![Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} = 5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}](https://tex.z-dn.net/?f=Q_%7B%5Ctext%7Bc%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BHI%5D%7D%5E%7B2%7D%7D%20%7B%5Ctext%7B%5BH%24_%7B2%7D%24%5D%5BI%24_%7B2%7D%24%5D%7D%7D%20%3D%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.30%20%5Ctimes%200.15%7D%20%3D%20%205.56%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%200.30%20%5Ctimes%200.15%20%5Ctimes%205.56%20%3D%200.250%5C%5Cx%20%3D%20%5Csqrt%7B0.250%7D%20%3D%20%5Ctextbf%7B0.50%20mol%2FL%7D%5C%5C%5Ctext%7BThe%20initial%20concentration%20of%20HI%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.50%20mol%2FL%7D%7D%24%7D)
3. Plot the initial points
The graph below shows the initial concentrations plotted on the vertical axis.
