Answer:
The element that is oxidized is carbon.
Its oxidation state increased. It increased from -4 to +4
Explanation:
Oxidation is a process that involves increase in oxidation number.
The oxidation number of carbon in CH4 is -4
C + (1×4) = 0
C + 4 = 0
C = 0 - 4 = -4
The oxidation number of carbon in CO2 is +4
C + (2×-2) = 0
C - 4 = 0
C = 0+4 = 4
Increase in the oxidation number of carbon from -4 to +4 means carbon is oxidized
Answer:
6.4 m/s
Explanation:
From the equation of continuity
A1V1=A2V2 where A1 and V1 are area and velocity of inlet respectively while A2 and V2 are the area and velocity of outlet respectively
where r1 and r2 are radius of inlet and outlet respectively
v1 is given as 1.6 m/s
Therefore
Answer:
Tech A is correct
Explanation:
Tech A is right as its V- angle is identified by splitting the No by 720 °. Of the piston at the edge of the piston.
Tech B is incorrect, as the V-Angle will be 720/10 = 72 for the V-10 motor, and he says 60 °.
Answer:
At the point when the quantity of bit strings is not exactly the quantity of processors, at that point a portion of the processors would stay inert since the scheduler maps just part strings to processors and not client level strings to processors. At the point when the quantity of part strings is actually equivalent to the quantity of processors, at that point it is conceivable that the entirety of the processors may be used all the while. Be that as it may, when a part string obstructs inside the portion (because of a page flaw or while summoning framework calls), the comparing processor would stay inert. When there are more portion strings than processors, a blocked piece string could be swapped out for another bit string that is prepared to execute, in this way expanding the use of the multiprocessor system.When the quantity of part strings is not exactly the quantity of processors, at that point a portion of the processors would stay inert since the scheduler maps just bit strings to processors and not client level strings to processors. At the point when the quantity of bit strings is actually equivalent to the quantity of processors, at that point it is conceivable that the entirety of the processors may be used at the same time. Be that as it may, when a part string hinders inside the piece (because of a page flaw or while summoning framework calls), the relating processor would stay inert. When there are more portion strings than processors, a blocked piece string could be swapped out for another bit string that is prepared to execute, along these lines expanding the usage of the multiprocessor framework.
