mole is the standardized form of molarity
Answer:
12.0108408
Explanation:
Denote the element with a letter like say X. Since it has a subscript of 5, then, X5.
Molecular mass=102.133g/mol.
% of X in compound =58.8/100
=0.588
Mass of X in the compound = 0.588*102.133 ( the % of X in compound * molar mass of compound)
= 60.054204
X5=60.054204
Then element X has a mass of 60.054204/5=12.0108408
Answer:
[H₂] = 1.61x10⁻³ M
Explanation:
2H₂S(g) ⇋ 2H₂(g) + S₂(g)
Kc = 9.30x10⁻⁸ = ![\frac{[H_{2}]^2[S_{2}]}{[H_{2}S]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH_%7B2%7D%5D%5E2%5BS_%7B2%7D%5D%7D%7B%5BH_%7B2%7DS%5D%5E2%7D)
First we <u>calculate the initial concentration</u>:
0.45 molH₂S / 3.0L = 0.15 M
The concentrations at equilibrium would be:
[H₂S] = 0.15 - 2x
[H₂] = 2x
[S₂] = x
We <u>put the data in the Kc expression and solve for x</u>:
We make a simplification because x<<< 0.0225:
x = 8.058x10⁻⁴
[H₂] = 2*x = 1.61x10⁻³ M
Answer:
Molecular formula => C₃H₈O₃
Explanation:
From the question given above, the following data were obtained:
Carbon (C) = 39.12%
Hydrogen (H) = 8.75%
Oxygen (O) = 51.12%
Molar mass of compound = 92.09 g/mol
Molecular formula =?
Next, we shall determine the empirical formula of the compound. This can be obtained as follow:
C = 39.12%
H = 8.75%
O = 51.12%
Divide by their molar mass
C = 39.12 / 12 = 3.26
H = 8.75 / 1 = 8.75
O = 51.12 / 16 = 3.195
Divide by the smallest
C = 3.26 / 3.195 = 1
H = 8.75 / 3.195 = 2.7
O = 3.195 / 3.195 = 1
Thus, the empirical formula is CH₂.₇O
Finally, we shall determine the molecular formula of the compound. This can be obtained as follow:
Empirical formula = CH₂.₇O
Molar mass of compound = 92.09 g/mol
Molecular formula =?
Molecular formula = Empirical formula × n
Molecular formula = [CH₂.₇O]ₙ
92.09 = [12 + (2.7×1) + 16] × n
92.09 = 30.7n
Divide both side by 30.7
n = 92.09 / 30.7
n = 3
Molecular formula = [CH₂.₇O]ₙ
Molecular formula = [CH₂.₇O]₃
Molecular formula = C₃H₈O₃
