Answer:
The temperature would be reduced by half
Explanation:
Charles' Law => V ∝ T => V = kT => k = V/T
For two sets of T vs V conditions, the system constant (k) remains unchanged and k₁ = k₂ => V₁/T₁ = V₂/T₂.
Therefore, if V₁ is reduced to 1/2V₁ = V₂ => V₁/T₁ = 1/2V₁/T₂ => V₁/T₁ = V₁/2T₂
and solving for T₂ => 1/T₁ = 1/2T₂ => 2T₂ = T₁ => T₂ = 1/2T₁
∴ The initial temperature (T₁) would be reduced by half or, T₂ = 1/2T₁
The concentrations : 0.15 M
pH=11.21
<h3>Further explanation</h3>
The ionization of ammonia in water :
NH₃+H₂O⇒NH₄OH
NH₃+H₂O⇒NH₄⁺ + OH⁻
The concentrations of all species present in the solution = 0.15 M
Kb=1.8 x 10⁻⁵
M=0.15
![\tt [OH^-]=\sqrt{Kb.M}\\\\(OH^-]=\sqrt{1.8\times 10^{-5}\times 0.15}\\\\(OH^-]=\sqrt{2.7\times 10^{-6}}=1.64\times 10^{-3}](https://tex.z-dn.net/?f=%5Ctt%20%5BOH%5E-%5D%3D%5Csqrt%7BKb.M%7D%5C%5C%5C%5C%28OH%5E-%5D%3D%5Csqrt%7B1.8%5Ctimes%2010%5E%7B-5%7D%5Ctimes%200.15%7D%5C%5C%5C%5C%28OH%5E-%5D%3D%5Csqrt%7B2.7%5Ctimes%2010%5E%7B-6%7D%7D%3D1.64%5Ctimes%2010%5E%7B-3%7D)
![\tt pOH=-log[OH^-]\\\\pOH=3-log~1.64=2.79\\\\pH=14-2.79=11.21](https://tex.z-dn.net/?f=%5Ctt%20pOH%3D-log%5BOH%5E-%5D%5C%5C%5C%5CpOH%3D3-log~1.64%3D2.79%5C%5C%5C%5CpH%3D14-2.79%3D11.21)
Answer:
The correct answer is - sulfur.
Explanation:
In the periodic table, there are 18 groups and 7 rows or periods arranged according to their atomic number or electronic configuration. In the question, it is mentioned that the desired element atomic mass is less than the atomic mass of the selenium which is 78.96, and more than oxygen which is 15.99 with 6 electron valence and present in the third row.
As it has 6 valency of electron it must be in the 16 group of the table that comprises the 6 valency and as it is located in the 3rd row it must be sulfur that also has an atomic mass between selenium and oxygen.
