Answer:
The current temperature is -15° (15° below zero)
Explanation:
The temperature drops 10°:
T-10°
It will reach 25° below zero:
T - 10° = -25°
We add 10° in both members of the equation:
T - 10° +10° = -25° +10°
The equation is simplified as follows:
T = -25° +10°
T = -15°
The integer -15° can be expressed as 15° below zero.
Answer:
Fluoride is commonly used in dentistry to strengthen enamel, which is the outer layer of your teeth. Fluoride helps to prevent cavities. It’s also added in small amounts to public water supplies in the United States and in many other countries. This process is called water fluoridation.
Explanation:
Answer:
the correct awnser is b 1-heptene
Answer:
SN2
Explanation:
The first step of ether cleavage is the protonation of the ether since ROH is a better leaving group than RO-.
The second step of the reaction may proceed by either SN1 or SN2 mechanism depending on the structure of the ether. Methyl and primary ethers react with HI by SN2 mechanism while tertiary ethers react with HI by SN1 mechanism. Secondary ethers react with HI by a mixture of both mechanisms.
Dipentyl ether is a primary ether hence when treated with HI, the reaction with HI proceeds by SN2 mechanism as explained above.
Answer:the pH is 12
Explanation:
First We need to understand the structure of trimethylamine
Due to the grades of the bond in the nitrogen with a hybridization sp3 is 108° approximately, then is generated a dipole magnetic at the upper side of the nitrogen, this dipole magnetic going to attract a hydrogen molecule of the water making the water more alkaline
C3H9N+ H2O --> C3H9NH + OH-
![k=\frac{[C3H9NH]*[OH-]}{[C3H9N]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B%5BC3H9NH%5D%2A%5BOH-%5D%7D%7B%5BC3H9N%5D%7D)
Then:
The concentration of the trimethylamine is 0.3 and the concentration of the ion C3H9NH is equal to the OH- relying on the stoichiometric equation. We could find the concentration of the OH- ion with the square root of the multiplication between k and the concentration of trimethylamine
[OH-]=
[OH-]=0.01
pH=14-(-log[OH-])
pH=12
