Enormous O unpredictability is in reference to the most exceedingly terrible conceivable development rate of the calculation. So O(N log N) implies that it will never keep running in some time more terrible than O(N log N). So in spite of the fact that Al's calculation scales superior to Bob's quadratic algo, it doesn't really mean it is better for ALL info sizes.
Maybe there is critical overhead in building up it, for example, making a lot of clusters or factors. Remember that even an O(N log N) calculation could have 1000 non settled circles that official at O(N) and still be viewed as O(N log N) the length of it is the most exceedingly awful part.
Answer : The enthalpy of the reaction = -1839.6 KJ
Solution : Given,
= -520.0 KJ/mole
= -1699.8 KJ/mole
The balanced chemical reaction is,

Formula used :


We know that the standard enthalpy of formation of the element is equal to Zero.
Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.
Now, put all the values in above formula, we get
![\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]](https://tex.z-dn.net/?f=%5CDelta%20%28H_%7Bf%7D%29_%7Breaction%7D%3D%5B2moles%5Ctimes%20%28-1699.8%20KJ%2Fmole%29%7D%2B3moles%5Ctimes%20%280%5Ctext%7B%20KJ%2Fmole%7D%7D%29%5D-%5B%283moles%5Ctimes%28-520.0KJ%2Fmole%20%7D%2B4moles%5Ctimes%280%5Ctext%7B%20KJ%2Fmole%7D%29%5D)
= (-3399.6) + (1560)
= -1839.6 KJ
There are 100 degrees between the freezing (0°) and boiling points (100°) of water on the Celsius scale and 180 degrees between the similar points (32° and 212°) on the Fahrenheit scale.
Problem: Two scientists are doing an experiment designed to identify the boiling point
Answer: 250°F is the higher temperature by 2°F
Answer:
Mass of oxygen in glucose = 29.3g
Explanation:
Mass of glucose given is 55grams.
We are to find the mass of oxygen in this compound.
In the compound we have 6 atoms of oxygen.
Solution
To find the mass of oxygen in glucose, we calculate the formula mass of glucose. We now divide the formula mass of the oxygen atom with that of the glucose and multiply by the given mass to find the unkown mass.
Atomic mass of C = 12g
H = 1g
O = 16g
Formula mass of C₆H₁₂O₆ = {(12x6) + (1x12) + (16x6)} = 180
Mass of O in glucose =
x 55
=
x 55
= 0.53 x 55
Mass of oxygen in glucose = 29.3g