We need (i) the stoichiometric equation, and (ii) the equivalent mass of dihydrogen.
Explanation:
1
2
N
2
(
g
)
+
3
2
H
2
(
g
)
→
N
H
3
(
g
)
11.27
g
of ammonia represents
11.27
⋅
g
17.03
⋅
g
⋅
m
o
l
−
1
=
?
?
m
o
l
.
Whatever this molar quantity is, it is clear from the stoichiometry of the reaction that 3/2 equiv of dihydrogen gas were required. How much dinitrogen gas was required?
Adopting the number of avogrado 6.02 * 10²³ / mol
<span>Sodium chloride (table salt)</span> Molar Mass = 58.44 g / mol
We will first have to find the number of moles in 35 grams of the element, like this:
1 mol ----------------- 58.44 g
X ---------------------- 35 g
58.44 * x = 35 * 1
58.44x = 35

X = 0.598904...
X ≈ 0.60<span> mol </span>
Now we will find how many atoms there are in 0.60 mol of this element, like this:
1 mol -------------------- 6.02 * 10²³ atoms
0.60 mol ----------------- X
X = 0.60 * 6.02 * 10²³
Answer:
there are no examples but 1 example is H2O which has 2 elements combining a compound.
Explanation:
The value of Kc for the thermal decomposition of H₂S is 2.2 x 10⁻⁴ at 1400 K:
2 H₂S(g) ↔ 2 H₂(g) + S₂(g)
initial 3.5 M 0 0
at equilibrium 3.5 M - 2x 2x x
Kc = [S₂][H₂]² / [H₂S]²
2.2 X 10⁻⁴ = x(2x)² / (3.5 - 2x)²
2.2 x 10⁻⁴ = 4 x³ / (3.5)² Assuming x <<<<< 3.5
x = 0.088
Thus [H₂S] = 3.324 M
Toulene = 35.6 g
Benzene = 125 g = 0.125 kg
Molecular weight of Toluene C6H5CH3 = 92.1g/mol
Moles of toulene = 35.6 g / 92.1 g/mol = 0.3865 mol
Now the molarity of the toulene in the given solution = 0.3865 / 0.125 = 3.092 m
Molarity of C6H5CH3 = 3.092 m