Using the formula:
F = k.q1.q2/r2
Manipulating for r.
<span>r = √(Kq1q2/F)
r = </span><span>1.5 x 10^-2 meters</span>
Answer:
In the transition series, atomic size across a period decreases at first but then remains relatively constant.
The transition elements in a period show a steady increase in electronegativity.
Explanation:
In considering the transition series, we observe that atomic sizes of the elements decreases first and subsequently remain constant. The reason for the initial decrease in atomic size is the increase in nuclear charge across the period. After the first few elements in the period, the atomic size remains relatively constant due to shielding effect of the inner d electrons which opposes the increase in effective nuclear charge.
It is also observed that electro negativity increases smoothly across the period for the transition series. Consequently, the transition series become less electro positive across the period.
Answer:
Air temperature is not accurate but it can measure by the air is cold or hot
Hope this will help you
Answer:
A.
Explanation:
The β-sheet is a twisted pattern in which the protein strands are laterally linked through hydrogen bonds. These β-sheet motifs are segments of 3 to 10 amino acids that are specially configured to form beta antiparallel strands. In β-sheet regions, the hydrogen bonds are localized among carbonyl and amino groups of the polypeptide backbone, and side chains (i.e., the R groups) are extended up and down in the plane of the β-sheet.
<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 />
