Answer:
<h2>
<em>Option</em><em> </em><em>C</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>correct </em><em>option</em><em>.</em></h2>
<em>The</em><em> </em><em>Valence </em><em>electrons</em><em> </em><em>in</em><em> </em><em>S</em>
<em>Expla</em><em>nation</em><em>:</em>
<em>Sulphur</em><em> </em><em>has</em><em> </em><em>highest</em><em> </em><em>effect</em><em>ive</em><em> </em><em>nuclear</em><em> </em><em>charge</em><em> </em><em>as</em><em> </em><em>in </em><em>a</em><em> </em><em>period</em><em> </em><em>there</em><em> </em><em>is </em><em>increase</em><em> </em><em>in</em><em> </em><em>effective</em><em> </em><em>nuclear</em><em> </em><em>charge</em><em> </em><em>from</em><em> </em><em>left </em><em>to</em><em> </em><em>right.Sulphur</em><em> </em><em>is </em><em>in </em><em>the</em><em> </em><em>end </em><em>in</em><em> </em><em>a</em><em> </em><em>period</em><em> </em><em>leaving</em><em> </em><em>Mg</em><em> </em><em>and</em><em> </em><em>Al</em><em> </em><em>in</em><em> </em><em>left </em><em>side.</em>
<em>Hope</em><em> </em><em>this</em><em> </em><em>helps.</em><em>.</em><em>.</em><em>.</em>
<em>Good</em><em> </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em><em>.</em>
Answer:
12.02 g
Explanation:
From the question given above, the following data were obtained:
Half life (t½) = 2 days
Original amount (N₀) = 96 g
Time (t) = 6 days
Amount remaining (N) =..?
Next, we shall determine the rate of disintegration of the isotope. This can be obtained as follow:
Half life (t½) = 2 days
Decay constant (K) =?
K = 0.693 / t½
K = 0.693 / 2
K = 0.3465 /day
Therefore, the rate of disintegration of the isotope is 0.3465 /day.
Finally, we shall determine the amount of the isotope remaining after 6 days as follow:
Original amount (N₀) = 96 g
Time (t) = 6 days
Decay constant (K) = 0.3465 /day.
Amount remaining (N) =.?
Log (N₀/N) = kt / 2.303
Log (96/N) = (0.3465 × 6) / 2.303
Log (96/N) = 2.079/2.303
Log (96/N) = 0.9027
Take the anti log of 0.9027
96/N = anti log (0.9027)
96/N = 7.99
Cross multiply
96 = N × 7.99
Divide both side by 7.99
N = 96 /7.99
N = 12.02 g
Therefore, the amount of the isotope remaining after 6 days is 12.02 g
Answer:
[OH-] = 1.0 x 10-10 M
Explanation:
The acidity of a solution can be determined directly from the concentration of the hydrogen ions and indirectly from the concentrations of the hydroxide ions.
Generally, for a neutral solution we have;
[H3O+] = [OH-] = 1.0 x 10-7 M
For an acidic solution;
[H3O+] > 1.0 x 10-7 M
[OH-] < 1.0 x 10-7 M
Comparing the options the correct option is;
[OH-] = 1.0 x 10-10 M
<span>Pitch is sometimes defined as the fundamental frequency of a sound wave (i.e. generally, the lowest frequency in a given sound wave). For most practical purposes, this is fine, and pitch and frequency can be thought of as equivalent. On the other hand, for most practical purposes, amplitude can be thought of as volume.However, technically, pitch (and volume) are human perceptions. Thus, our perception of pitch and volume are not solely based on frequency and amplitude respectively, but are based on a combination of both (and even other factors). Frequency overwhelming dictates perceived pitch, but amplitude also does have some small, small effect on our pitch perception, especially when it is very large. For example, a very loud sound can have a different <span>perceived </span>pitch than you would predict from its frequency alone.That all being said, usually these effects are negligible, and pitch can be thought of as equivalent to fundamental frequency.
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