Intermolecular force for solids is high. Whereas low in gases. The smell of agarbatti spreads immediately because the molecules of air diffuses very fastly.
Answer:
A. DH° = –36 kJ
Explanation:
It is possible to obtain DH° of a reaction by the sum of DH° of half reactions. The DH° of the reaction:
B₂H₆(g) → 2B(s) + 3H₂(g)
Could be obtained from:
<em>(1) </em>2B(s) + 1.5O₂(g) → B₂O₃(s) DH° = –1273kJ
<em>(2) </em>B₂H₆(g) + 3O₂(g) → B₂O₃(s) + 3H₂O(g) DH° = –2035kJ
<em>(3) </em>H₂(g) + 0.5O₂(g) → H₂O(g) DH° = –242kJ
The sum of (2) - (1) gives:
B₂H₆(g) + 1.5O₂(g) → 2B(s) + 3H₂O(g) DH° = -2035kJ - (-1273kJ) = -762kJ
Now, this reaction - 3×(3):
B₂H₆(g) → 2B(s) + 3H₂(g) DH° = -762kJ - (3×-242kJ) = -36kJ
Thus, right answer is:
<em>A. DH° = –36 kJ</em>
Answer:
True
Explanation:
Salt is used to keep ice from forming but it only helps with certain tempuratures is still able to freeze
Answer:
After 26.0s, the concentration of HI decreases from 0.310M to 0.0558M.
Explanation:
Based on the reaction of the problem, you have as general kinetic law for a first-order reaction:
ln[HI] = -kt + ln [HI]₀
<em>Where [HI] is actual concentration after time t, </em>
<em>k is rate constant </em>
<em>and [HI]₀ is initial concentration of the reactant.
</em>
Initial concentration of HI is 0.310M,
K is 0.0660s⁻¹,
And the actual concentration is 0.0558M:
ln[HI] = -kt + ln [HI]₀
ln[0.0558M] = -0.0660s⁻¹*t + ln [
0.310M]
-1.7148 = -0.0660s⁻¹*t
26.0s = t
<h3>After 26.0s, the concentration of HI decreases from 0.310M to 0.0558M</h3>
<em />
Answer:
- <u><em>Option A.) 2 ppm</em></u>
Explanation:
<u><em>60 particles in 30 million</em></u> can be expressed as a fraction:
And, dividing both numerator and denominator by 30, that can be simplified as:
That means 2 particles of sulfur dioxide in 1 million particles of sample, which is 2 parts per million or<u> 2 ppm</u>, which is the option A.
