So what we know:
-Atomic Mass = Protons + Neutrons
-Atomic Number is the number of protons
Magnesium's atomic number is 12, so the natural occurring isotope for magnesium is Mg-12 (12 protons and 12 neutrons). Added up we have an atomic mass of 24 amu. Which means if we added one neutron in Mg-13, our atomic mass would be 25 amu.
We can use the equation:
(amu of isotope 1)x + (amu of isotop 2)(x-1) = Average atomic mass
where isotope 1 is the fractional abundance we're solving for.
Plugged in it looks like this:
24x + 25(1-x) = 24.3
Now to solve for x:
24x + 25 - 25x = 24.3
-x + 25 = 24.3
-x = -.7
x = .7
So in this case, the fractional abundance of Mg-12 would be .7, or 70%.<span />
Answer:
the answer is 0.4588162459
Explanation:
1 mole = 0.010195916576195
= 45 ×"
Answer:
Explanation:
<u>1) Data:</u>
a) Hypochlorous acid = HClO
b) [HClO} = 0.015
c) pH = 4.64
d) pKa = ?
<u>2) Strategy:</u>
With the pH calculate [H₃O⁺], then use the equilibrium equation to calculate the equilibrium constant, Ka, and finally calculate pKa from the definition.
<u>3) Solution:</u>
a) pH
b) Equilibrium equation: HClO (aq) ⇄ ClO⁻ (aq) + H₃O⁺ (aq)
c) Equilibrium constant: Ka = [ClO⁻] [H₃O⁺] / [HClO]
d) From the stoichiometry: [CLO⁻] = [H₃O⁺] = 2.29 × 10 ⁻⁵ M
e) By substitution: Ka = (2.29 × 10 ⁻⁵ M)² / 0.015M = 3.50 × 10⁻⁸ M
f) By definition: pKa = - log Ka = - log (3.50 × 10 ⁻⁸) = 7.46
Answer:
The percent yield of this reaction is 70%
Explanation:
The reaction is: N₂ + 3H₂ → 2NH₃
We only have the mass of H₂, so we assume that N₂ is in excess
We convert the mass to moles, to work with the reaction:
450 g . 1mol / 2 g = 225 moles
Ratio is 2:3. 3 moles of H₂ can produce 2 moles of ammonia
Therefore 225 moles of H₂ will produce (225 .2)/ 3 = 150 moles
This is the 100% yield reaction → We convert the moles of NH₃ to mass
150 mol . 17g /1mol = 2550 g
Percent yield = (Produced yield/Theoretical yield) .100
Percent yield = (1575g/2550g) . 100 = 70%