The modern periodic table is organized according to atomic mass true.
Answer : The correct option is, (C) 1.1
Solution : Given,
Initial moles of
= 1.0 mole
Initial volume of solution = 1.0 L
First we have to calculate the concentration
.


The given equilibrium reaction is,

Initially c 0
At equilibrium

The expression of
will be,
![K_c=\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)

where,
= degree of dissociation = 40 % = 0.4
Now put all the given values in the above expression, we get:



Therefore, the value of equilibrium constant for this reaction is, 1.1
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Kc = [CS2] / [S2][C] but, since C is solid, it doesn't fit here
<span>Kc = [CS2] / [S2] </span>
<span>S2(g) + C(s) ---> CS2(g) </span>
<span>13.8mole S2 / 6.05L = 2.28M </span>
<span>Kc = x / 2.28-x </span>
<span>9.4 x (2.28-x) = x </span>
<span>21.43 - 9.4x = x </span>
<span>21.42 = 10.4x </span>
<span>x = 2.05M = [CS2] </span>
<span>2.05M x 6.05L = 12.46moles </span>
<span>12.46moles x 76gmole = 946.96g CS2</span>
Answer:
Mass = 3.84 g
Explanation:
Given data:
Mass of hydrogen sulfide = 2.7 g
Mass of oxygen required = ?
Solution:
Chemical equation:
2H₂S + 3O₂ → 2H₂O + 2SO₂
Number of moles of hydrogen sulfide:
Number of moles = mass/ molar mass
Number of moles = 2.7 g / 34 g/mol
Number of moles = 0.08 mol
Now we will compare the moles of hydrogen sulfide with oxygen.
H₂S : O₂
2 : 3
0.08 : 3/2 ×0.08 = 0.12 mol
Mass of oxygen;
Mass = number of moles × molar mass
Mass = 0.12 mol × 32 g/mol
Mass = 3.84 g