Answer: 13.42g
Explanation:
1mole of carbon =12g
1mole (12g) of carbon contains 6.02x10^23 atoms.
Therefore, Xg of carbon will contain 6.73x10^23 atoms i.e
Xg of carbon = (12x6.73x10^23)/6.02x10^23 = 13.42g
Caffeine is not optically active because it is a planar heterocycle. When you say optically active, it means that compounds have molecules which are chiral. Chirality is the property of molecules as a result of its structure. And when you say planar heterocycle, molecules are flat and the ring should contain atoms located in a similar plane.
Answer:
A. PO43-
Explanation:
in the given is buffer . so H2PO4- and HPO42- both are present with equal concentration . Na+ is spectator ion it is also present in the concentration higher than the given species above .
but PO4-3 is not present . so it is lowest concentration
Answer:
D.All three graph lines peak over the same few weeks of winter.
Explanation:
took the test
Answer:
There are 3 steps of this problem.
Explanation:
Step 1.
Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.
Step 2.
Enthalpy of saturated liquid Haq = 781.124 J/g
Enthalpy of saturated vapour Hvap = 2779.7 J/g
Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g
Step 3.
In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy
So, H1=H2
H2= (1-x)Haq+XHvap.........1
Putting the values in 1
2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}
= 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)
1904.976 (J/g) = x1998.576 (J/g)
x = 1904.976 (J/g)/1998.576 (J/g)
x = 0.953
So, the quality of the wet steam is 0.953
