Answer: Yes.
Explanation: Because of the crystalline structure.
8. How did the measured angular magnification of the telescope compare with the theoretical prediction?
Complete Question
The complete question is shown on the first uploaded image
Answer:
The theoretical angular magnification lies within the angular magnification range
Explanation:
From the question we are told that
The focal length of B is [tex]f_{objective } = 43.0 \ cm[/tex]
The focal length of A is [tex]f_{eye} = 10.4 \ cm[/tex]
The theoretical angular magnification is mathematically represented as
[tex]m = \frac{f_{objective }}{f_{eye}} = \frac{43.0}{10.4}[/tex]
[tex]m = \frac{f_{objective }}{f_{eye}} = 4.175[/tex]
Form the question the measured angular magnification ranges from 4 -5
So from the value calculated and the value given we can deduce that the theoretical angular magnification lies within the angular magnification range
The voltage for the following cell is +0.731 V. Find Kb for the organic base RNH2. Use E_SCE = 0.241V.
Pt(s)│H2 (1.00 atm)│RNH2 (0.100M), RNH3 + (0.0500M)║SCE
Answer: Kb = 1.89 × 10⁻⁶
Explanation:
R-NH₃+ (aq) <----------> R-NH₂ (aq) + H^+ (aq)
Ecell = - ( 0.0592 / n ) log Kₐ
Where, Ecell = 0.731 - 0.241 = 0.490 V
Therefore, 0.490 = - (0.0592 / 1 ) log Kₐ
Therefore, Kₐ = 5.30 × 10⁻⁹
Thus, Kb = Kw / Ka = ( 1.0 × 10⁻¹⁴ / 5.30 × 10⁹ ) = 1.89 × 10⁻⁶