ABO identification system is the system of determining the blood type of the Human being by checking the presence of Antigens "A" & "B" on the surface of their RBCs. If A is present, then it is A Blood group, if B, then it is B blood group, if both, then AB and if neither then it is called as O blood group.
Rh Factor is the type by which we classify individuals into two groups - Rh +ve with Rh factor on their surface and Rh - ve, which lacks of it
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Answer:
![[N_2]=0.0866M](https://tex.z-dn.net/?f=%5BN_2%5D%3D0.0866M)
Explanation:
Hello there!
In this case, in agreement to the chemical reaction, it is possible for us to figure out the equilibrium concentration of the N2 product, via an ICE table plugged in the equilibrium expression:
![Kc=\frac{[N_2][O_2]}{[NO]^2}\\\\2.4x10^3=\frac{x*x}{(0.175-2x)^2}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BN_2%5D%5BO_2%5D%7D%7B%5BNO%5D%5E2%7D%5C%5C%5C%5C2.4x10%5E3%3D%5Cfrac%7Bx%2Ax%7D%7B%280.175-2x%29%5E2%7D)
In such a way, when solving for x via quadratic equation or just a solver, it is possible to obtain:
In such a way, since the root 0.0884 M produce a negative concentration of NO (0.175-2*0.0884=-0.0018M), we infer that the correct root is 0.0866 M; therefore, the concentration of N2 at equilibrium is equal to x:
![[N_2]=x=0.0866M](https://tex.z-dn.net/?f=%5BN_2%5D%3Dx%3D0.0866M)
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Answer:
0.13 g
Explanation:
mass of aluminum required = ( Dislocation length) / ( Dislocation density) × (density of metal)
3000 miles to cm ( 1 mile = 160934 cm) = 3000 miles × 160934 cm / 1 mile = 482802000 cm
density of Aluminium = 2.7 g /cm³
dislocation density of aluminum = 10¹⁰ cm³
mass of aluminum required = (482802000 cm × 2.7 g/cm³) / 10¹⁰ cm³ = 0.13 g
<span>8.21 L of C3H8(g)
Lets take c as the molar volume at that temperature.
c L <><> 5c L
C3H8 (g) + 5O2 (g) --> 3CO2 + 4H2O + Q
8.21 L <><> x L
x = (8.21 * 5c)/c = 8.21 * 5 = 41.05 L O2 consumed for a 100% yield.</span>
It would be atomic masses of the same atoms, and that atoms will be isotopes.
