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Yes, It is important to know the volume of Unknown acid or base to be titrated.
Titration is carried out in order to find out the concentration (i.e. molarity) of unknown acid or base. In this process a standard solution of acid or base is taken and is titrated with known volume of of titrant. At end point (neutralization) the amount of standard titrant utilized is calculated and following formula is employed to calculate the unknown concentration of unknown solution.
M₁V₁/n₁ = M₂V₂/n₂
Given:
Moles of H2 = 0.300
Moles of I2 = 0.400
Moles of HI = 0.200
Keq = 870
To determine:
Amounts of the mixture at equilibrium
Explanation:
H2(g) + I2(g) ↔ 2HI(g)
Initial 0.3 0.4 0.2
Change -x -x +2x
Eq (0.3-x) (0.4-x) (0.2+2x)
Keq = [HI]²/[H2][I2]
870 = (0.2+2x)²/(0.3-x)(0.4-x)
x = 0.29 moles
Amounts at equilibrium:
[HI] = 0.2 + 2(0.29) = 0.78 moles
[H2] = 0.3-0.29 = 0.01 moles
[I2] = 0.4-0.29 = 0.11 moles
<em>The number of atomic orbitals is the same for all elements in a period. Every element in the top row (the first period), for example, has only one orbital for its electrons. The electrons of all the elements in the second row (second period) have two orbitals. Every row adds an orbital as you progress down the table.</em>
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<em>I hope this helps <3</em>
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Answer: The pressure of nitrogen will be 2.0 atm
Explanation:
According to ideal gas equation:
P = pressure of gas = 1.0 atm
V = Volume of gas = 10.0 L
n = number of moles = ?
R = gas constant =
T =temperature =
Now again using ideal gas equation :
![P=\frac{nRT}{V}=\frac{0.41mol\times 0.0821Latm/Kmol\times 297K}{5.0L}=2.0atm](https://tex.z-dn.net/?f=P%3D%5Cfrac%7BnRT%7D%7BV%7D%3D%5Cfrac%7B0.41mol%5Ctimes%200.0821Latm%2FKmol%5Ctimes%20297K%7D%7B5.0L%7D%3D2.0atm)
Thus pressure of nitrogen will be 2.0 atm