PH + poH = 14
6.2 +poH = 14
poH = 7.8
The particles are quite tightly packed together but still have enough room to be able to move and flow, their bonds aren't as strong as a solids are
Answer:
Add Iodine-KI reagent to a solution or directly on a potato or other materials such as bread, crackers, or flour. A blue-black color results if starch is present. If starch amylose is not present, then the color will stay orange or yellow
<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 />
Explanation:
Moles of N2 = 35.0g / (28g/mol) = 1.25mol
Moles of H2 = 60.0g / (2g/mol) = 30.0mol
Since 1.25mol * 3 < 30.0mol, nitrogen is limiting.
Moles of NH3 = 1.25mol * 2 = 2.50mol.
Mass of NH3 = 2.50mol * (17g/mol) = 42.5g.
30.0mol - 1.25mol * 3 = 26.25mol.
Excess mass of H2
= 26.25mol * (2g/mol) = 52.5g.
