The sum total<span> of the genetically based </span>variety<span> of </span>living organisms<span> in the </span>biosphere<span> is called a. species diversity. c. biodiversity.</span>
<span>Answer: 0.094%
</span><span>Explanation:
</span>
<span></span><span /><span>
1) Equilibrium chemical equation:
</span><span />
<span>Only the ionization of the formic acid is the important part.
</span><span />
<span>HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq).
</span><span />
<span>2) Mass balance:
</span><span />
<span> HCOOH(aq) HCOO⁻(aq) H⁺(aq).
Start 0.311 0.189
Reaction - x +x +x
Final 0.311 - x 0.189 + x x
3) Acid constant equation:
</span><span />
<span>Ka = [HCOO-] [N+] / [HCOOH] = (0.189 + x) x / (0.311 -x)
</span><span />
<span>= (0.189 + x )x / (0.311 - x) = 0.000177
4) Solve the equation:
You can solve it exactly (it will lead to a quadratic equation so you can use the quadratiic formula). I suggest to use the fact that x is much much smaller than 0.189 and 0.311.
</span><span />
<span>With that approximation the equation to solve becomes:
</span><span>0.1890x / 0.311 = 0.000177, which leads to:</span>
<span /><span>
x = 0.000177 x 0.311 / 0.189 = 2.91 x 10⁻⁴ M
5) With that number, the percent of ionization (alfa) is:
</span><span />
<span>percent of ionization = (moles ionized / initial moles) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (concentration of ions / initial concentration) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (0.000291 / 0.311)x 100 = 0.0936% = 0.094%
</span>
<span></span><span />
Answer:
46.761g/mol
Explanation:
Given parameters:
Element = Hilarium , Hi
Isotopes: Hi- 45, Hi-46 and Hi- 48
Natural abundance of Hi-45 = 18.3%
Hi-46 = 34.5%
Hi-48 = 47.2%
Unknown:
Atomic weight of naturally occurring Hilarium = ?
Solution:
Isotopes have been studied extensively by mass spectrometry. The method is used to determine the proportion/percentage/fraction by which each of the isotopes of an element occurs in nature. The proportion is called geonormal abundance. From this we can calculate the atomic weight of an element.
We can use the expression below to find this value:
Atomic weight = m₄₅α₄₅ + m₄₆α₄₆ + m₄₈α₄₈
m is the atomic mass of each isotope and α is the abundance
Atomic weight = (45 x ) + (46 x ) + (48 x )
Atomic weight of Hi = 8.235 + 15.870 + 22.656 = 46.761g/mol
The atomic radius decreases
Decomposition reaction /cracking reactions