Answer:
Change in internal energy (ΔU) = -9 KJ
Explanation:
Given:
q = –8 kJ [Heat removed]
w = –1 kJ [Work done]
Find:
Change in internal energy (ΔU)
Computation:
Change in internal energy (ΔU) = q + w
Change in internal energy (ΔU) = -8 KJ + (-1 KJ)
Change in internal energy (ΔU) = -8 KJ - 1 KJ
Change in internal energy (ΔU) = -9 KJ
Answer:
Nitrogen atom is small in size than phosphorus thus the bond between its nucleus and valence electrons are stronger than phosphorus.
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Answer:
Explanation:
The correct measurement is .710 s which is equal to .71 s so second measurement that is 0.71 s is most accurate measurement .
B. 0.71s is the most accurate .
Precision depends upon measuring instrument . Measurement by highly precise instrument has greater precision .
The measurement of 0.75 ± 0.002s must have been taken from high precise instrument because it is capable of making measurement upto 3 decimal points .
Hence
C 0.75 ± 0.002s is most precise measurement .
Answer:
The effective nuclear charge for a valence electron in oxygen atom: 
Explanation:
Effective nuclear charge ![[Z_{eff}]](https://tex.z-dn.net/?f=%5BZ_%7Beff%7D%5D)
is the net nuclear charge experienced by the electron in a given atom. It is always less than the actual charge of the nucleus [Z], due to shielding by electrons in the inner shells.
<em>It is equal to the difference between the actual nuclear charge or the atomic number (Z) and the shielding constant (s). </em>
<u>For an oxygen atom</u>-
Electron configuration: (1s²) (2s² 2p⁴)
<em>The atomic number (actual nuclear charge): </em>Z = 8
The shielding constant (s) for a valence electron can be calculated by using the Slater's rules:
⇒ s = 5 × 0.35 + 2 × 0.85 = 1.75 + 1.7 = 3.45
<u><em>Therefore, the effective nuclear charge for a valence electron in oxygen atom is:</em></u>
<u>Therefore, the effective nuclear charge for a valence electron in oxygen atom:</u>
The complete question is: Match the following; Disulfide BondsA. Covalent interactions not found in all proteins.Peptide BondsB. Covalent interactions found in all proteins.Long-range interactionsC. Non-covalent interaction formed primarily on the interior of water-soluble proteins.Hydrophobic coreD. Covalent or non-covalent interactions formed b/w amino acid far from each other in primary structure.
The answer
Disulfide Bonds (Covalent interactions not found in all proteins.)
Peptide Bonds (Covalent interactions found in all proteins.)
Long-range interactions (Covalent or non-covalent interactions formed b/w amino acid far from each other in primary structure.)
Hydrophobic core. (Non-covalent interaction formed primarily on the interior of water-soluble proteins.)
Explanation:
The phrases have been given the right meaning enclosed in a parentheses. The should be matched accordingly as presented.
