The number of half -lives that has passed after 105 hours for krypton-79 that has half-life of 35 hours is calculated as below
if 1 half life = 35 hours
what about 105 hours = ? half-lives
= (1 half life x105 hours) /35 hours = 3 half-lives has passed after 105 hours
Answer: options B,D and F
Explanation:
Since redox reactions are those which involves both oxidation and reduction
In B , Cu is oxidized and S gets reduced
D, Na gets oxidized and hydrogen gets reduced
F, carbon gets oxidized and Oxygen gets reduced
In g, there is no change in oxidation no of s in both product and Reactants is same +4
Similarly in the case of Ag and Mg.
Answer:
Carbon - 14
Oxygen - 16
Nitrogen - 15
Sulphur - 16
Explanation:
The question above is related to the "Periodic Table of Elements" which shows the proper arrangement of elements in a table. Every element falls on a<em> group/family</em> within the table. Each group has its own number, and the table has a total of<u> 18 groups</u><em> (from left to right). </em>They are classified according to <em>similarities in their characteristics</em>. For example, group 1 is composed of <em>alkali metals</em> while group 2 is composed of<em> alkali earth metals</em>.
The entropy will increase in pressure if you increase the pressure on the system the volume decrease the energies of the particles are in a smaller space so they are less spread out
Answer:
5x10⁻⁶ = [HTeH₄O₆⁺]
Explanation:
The first dissociation equilibrium of the telluric acid in water is:
H₂TeH₄O₆ + H₂O ⇄ HTeH₄O₆⁺ + H₃O⁺
Using H-H equation for telluric acid:
<em>pH = pKa + log₁₀ [HTeH₄O₆⁺] / [H₂TeH₄O₆]</em>
pKa of telluric acid is -logKa1
pKa = -log 2.0x10⁻⁸
pKa = 7.699
As concentration of [H₂TeH₄O₆] is 0.25M, replacing in H-H equation:
3.00 = 7.699+ log₁₀ [HTeH₄O₆⁺] / [0.25M]
-4.699 = log₁₀ [HTeH₄O₆⁺] / [0.25M]
2x10⁻⁵ = [HTeH₄O₆⁺] / [0.25M]
<h3>5x10⁻⁶ = [HTeH₄O₆⁺]</h3>
