I would have to say no. There are metamorphic rocks, igneous rocks, and sedimentary rocks. Igneous rocks are rocks that have solidified from magma or lava upon cooling, like lava rocks. Sedimentary rocks are rocks from smaller sediments, such as sandstone being made from, well, sand. Metamorphic rocks are the result of preexisting rocks in a response to changes in the environment. This includes changes in pressure, air temperature, mechanical stress, as well as taking away or adding chemical components. Metamorphic can be from igneous, sedimentary, or any other metamorphic rocks. Hopefully this help (with the added bonus that I explained all three type of rocks.
Answer:
See explanation
Explanation:
We can convert cyclohexanol to cyclohexene in the presence of a strong acid such as sulfuric acid catalyst in a test tube at 60 oC by heating up the mixture to about 80 oC. This is a dehydration reaction so water is removed to yield the alkene. A drying agent is used to remove any trace amount of water left in the system. This overall reaction is endothermic.
Also, the reverse is the case when we want to carry out the hydration of cyclohexene to yield cyclohexanol. The overall reaction is exothermic and involves the addition of more water to the alkene and then cooling down the system to about 40 oC.
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:
![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)

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>
Answer:
1620 cm⁻¹·M⁻¹
Explanation:
<em>Beer's Law </em>states that:
A = ε * b * C
Where A is the absorbance, ε is the Molar Absorptivity coefficient, b is the path length, and C is the molar concentration.
<u>Use the given data</u>:
0.405 = ε * 0.5 cm * 5x10⁻⁴ M
And <u>solve for ε</u>:
ε = 1620 cm⁻¹·M⁻¹
So the Molar Absorptivity coefficient of Compound X at this 265 nm is 1620 cm⁻¹·M⁻¹.