Answer:
Bohr's model of the hydrogen atom is based on three postulates:
1) An electron moves around the nucleus in a circular orbit,
2) An electron's angular momentum in the orbit is quantised,
3) The change in an electron's energy as it makes a quantum jump from one orbit to another is always accompanied by the emission or absorption of a photon. Bohr's model is semi-classical because it combines the classical concept of electron orbit (postulate 1) with the new concept of quantisation ( postulates 2 and ).
Q= mcΔT
1623 = 33.69g x c x (110.8 - 29.4)
1623 = 2742.366 g•°C x c
c = 0.59j/g•°C
<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 />
1) HOBr stands for hypobromous acid. On reacting with water, products formed are OBr- and H3O+. Following reaction occurs during this process.
<span> HOBr + H2O </span>⇄<span> OBr- + H3O+
2) HOBr is a weak acid and have a lower value of dissociation constant (Ka ~ </span><span>2.3 X 10^–9). Hence, </span><span> large number of undissociated HOBr molecules are left in solution, when the reaction is completed/reaches equilibrium.</span>
