Answer:
24.309 g/mol
Explanation:
To get the atomic mass, all we have to do is calculate with the masses of the three isotope, the real quantity present, taking account of the percent and then, do a sum of these three values. Like a pondered media.
For the first isotope:
23.99 * (78.99/100) = 18.95 g/mol
For the second isotope:
24.99 * (10/100) = 2.499 g/mol
For the last isotope:
25.98 * (11.01/100) = 2.86 g/mol
Now, let's sum all three together
AW = 18.95 + 2.499 + 2.86
AW = 24.309 g/mol
Answer:
the equilibrium concentration of HF is 2.85 M
Option a) 2.85 M is the correct answer.
Explanation:
Given the data in the question;
H₂ + F₂ ⇄ 2HF
I 1.69 M 1.69 M 0
C -x -x +2x
E 1.69-x 1.69-x +2x
given that Kc = 115
Kc = [ HF ]² / [H₂][F₂]
we substitute
115 = [ 2x ]² / [ 1.69-x ][ 1.69-x ]
lets find the square root of both sides
10.7238 = 2x / [ 1.69-x ]
10.7238[ 1.69-x ] = 2x
18.123222 - 10.7238x = 2x
2x + 10.7238x = 18.123222
12.7238x = 18.123222
x = 18.123222 / 12.7238
x = 1.424356
Hence, equilibrium concentration of HF = 2x
that is;
HF = 2 × 1.424356
HF = 2.8487 ≈ 2.85 M
Therefore, the equilibrium concentration of HF is 2.85 M
Option a) 2.85 M is the correct answer.
First, we will get the number of moles:
one mole contains Avogadro's number of atoms. Therefore, to know the number of moles in <span>7.83 × 10^24 atoms, we will simply do cross multiplication as follows:
number of moles = (</span><span>7.83 × 10^24*1) / (6.022 * 10^23) = 13 moles
From the periodic table:
mass of one mole of helium = 4 grams
Therefore:
mass of 13 moles of helium = 13*4 = 52 grams</span>
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Explanation:
