When adjusted for any changes in δh and δs with temperature, the standard free energy change δg∘t at 2400 k is equal to 1.22×105j/mol, then the equilibrium constant at 2400 k is 2.21×10−3. The answer to the statement is 2.21×10−3.
Each element<span> can usually be classified as a metal or a non-metal based on their ... They are usually </span>dull<span>and therefore show no metallic </span>luster<span> and they do not reflect ... </span>Dull<span>, Brittle solids; Little or no metallic </span>luster<span>; </span>High<span> ionization energies; </span>High<span> ...</span>
The book value of a machine, as shown on the balance sheet, is not relevant in a decision concerning the replacement of that machine by another machine: TRUE
<h3>
What is the book value?</h3>
- Book value is the worth of an asset based on its balance sheet account balance in accounting.
- The value of an asset is determined by subtracting the asset's original cost from any depreciation, amortization, or impairment expenses.
- Traditionally, a company's book value is equal to its total assets minus intangible assets and liabilities.
- In practice, however, depending on the source of the computation, book value may include either goodwill or intangible assets, or both.
- The value inherent in its employees, which is part of a company's intellectual capital, is always overlooked.
- When intangible assets and goodwill are specifically omitted, the indicator is frequently defined as "tangible book value."
Therefore, the statement "the book value of a machine, as shown on the balance sheet, is not relevant in a decision concerning the replacement of that machine by another machine" is TRUE.
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Complete question:
The book value of a machine, as shown on the balance sheet, is not relevant in a decision concerning the replacement of that machine by another machine. (Ignore taxes.) TRUE or FALSE
Answer:
The Answer should be Endothermic reaction
Explanation:
Reactions in which energy is absorbed are called Endothermic reactions.
Answer:
B and C
Explanation:
When we have to do a buffer solution we always have to choose the reaction that has the <u>pKa closer to the desired pH value</u>. When we find the pKa values we will obtain:
![pKa_1=-Log[6.9x10^-^3]=2.16](https://tex.z-dn.net/?f=pKa_1%3D-Log%5B6.9x10%5E-%5E3%5D%3D2.16)
![pKa_2=-Log[6.2x10^-^8]=7.20](https://tex.z-dn.net/?f=pKa_2%3D-Log%5B6.2x10%5E-%5E8%5D%3D7.20)
![pKa_3=-Log[4.8x10^-^13]=12.31](https://tex.z-dn.net/?f=pKa_3%3D-Log%5B4.8x10%5E-%5E13%5D%3D12.31)
The closer value is pKa2 with a value of 7.2. Therefore we have to use the second reaction. In which
is the <u>acid</u> and
is the <u>base</u>. Therefore the answer for the first question is B and the answer for the second question is C.