The North American plate is moving towards the west-southwest at about 2.3 centimeters every year mediated by the Mid-Atlantic Ridge, the spreading center, which gave rise to the Atlantic Ocean. The small Juan De Fuca plate, moving east-northeast at 4 centimeters every year, was once a component of much greater oceanic plates known as the Farallon plate.
The Farallon plate used to comprise what is now the Cocos plate of Mexico and Central America, and the Juan de Fuca plate in the region from N. Vancouver Island to the Cape Mendicino California, and a big sea floor tract in between. However, the middle portion of the Old Farallon plate disappeared underneath North America, it was subducted underneath California leaving the San Andreas fault system behind as the contact between the Pacific plates and North America.
The Juan De Fuca plate is still actively subducting underneath North America. Its movement is not smooth, however, rather sticky. The buildup of strain takes place until the fault dissociates and a few meters of Juan De Fuca get slid underneath North America in a big earthquake.
Answer:
They form a covalent bond
Answer: Mechanical digestion from chewing, chemical digestion from saliva
Explanation:
Answer:
Explanation has been given below
Explanation:
- In diaxial conformation of cis-1,3-disubstituted cyclohexane, 4 gauche-butane interactions along with syn-diaxial interaction are present. Hence it readily gets converted to diequitorial conformation where no such gauche-butane interaction is present
- In two possible conformations of trans-1,3-disubstituted cyclohexane, 2 gauche-butane interactions are present in each of them.
- Hence cis-1,3-disubstituted cyclohexane exists almost exclusively in diequitorial form. But trans-1,3-disubstituted cyclohexane has no such option.
- Trans-1,3-disubstituted cyclohexane experiences gauche butane interaction in each of the two conformations.
- Therefore cis-1,3-disubstituted cyclohexane is more stable than trans conformation
First, we will get the "n", the number of half lives, it is the elapsed time over the half life. In the problem, the time is measured in days, so we have
6/2 = 3
to get the ending amount of radioactive sample, we have
32g x (1/2)³ = 4 grams of radioactive sample ⇒ the amount left after 6 days
