Answer : Methane 
does not has any resonating structures or isomers.
Explanation : Methane has tetrahedral geometry where all the valence bonds of carbon are satisfied with hydrogen atoms which are similar to each other. They also do not produce any lewis dot structure. Therefore, methane cannot have resonating structures.
Answer:
A) Alkaline Earth Metals
Explanation:
Area Marked A would be Alkali Metals
Area Marked B would be Alkaline Earth Metals
Area Marked C would be Transition Metals
Area marked H would be Halogens
Answer:
b) O2, because it has weaker intermolecular forces
Explanation:
The preassure is produced by the collisions of the gas molecules with the walls of its container.
When the intermolecular forces between the gas molecules increase, those molecules start to "slow down" by effect of the interactions. The collisions decrease in frequency and intensity producing a smaller preassure in the container.
Both O2 and Cl2 are non-polar gases and the only intermolecular forces they have are the London ones. Given that the O2 molecules are smaller than the Cl2, the last ones attract each other with more strengh.
Being all that said, the container with the oxygen is expected to have a higher preassure.
It is none of the three. Conduction occurs when a heat source directly is touching an object.
Explanation: Number 1 is convection. Key term: convection currents. Number 2 is radiation, as radiation can pass through space. Number 3 is just explaining a part of the convection current.
Answer:
The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.
Explanation:
Consider the ICE take for the solubility of the solid, CuF₂ as:
CuF₂ ⇄ Cu²⁺ + 2F⁻
At t=0 x - -
At t =equilibrium (x-s) s 2s
The expression for Solubility product for CuF₂ is:
![K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5Cleft%20%5B%20Cu%5E%7B2%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20F%5E-%20%5Cright%20%5D%5E2)
Given s = 7.4×10⁻³ M
So, Ksp is:
Ksp = 1.6209×10⁻⁶
Now, we have to calculate the solubility of CuF₂ in NaF.
Thus, NaF already contain 0.20 M F⁻ ions
Consider the ICE take for the solubility of the solid, CuF₂ in NaFas:
CuF₂ ⇄ Cu²⁺ + 2F⁻
At t=0 x - 0.20
At t =equilibrium (x-s') s' 0.20+2s'
The expression for Solubility product for CuF₂ is:
Solving for s', we get
<u>s' = 4.0×10⁻⁵ M</u>
<u>The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.</u>
