平成18年度弁理士試験本試験問題とその傾向
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11 m n O A ABCDEFG A XML Wiki Ca Mg HCO Ca Mg moll ph moll e A A e B B AB T PAPB PAPB iiii i G A B A B ii C D D H iii E E RMulliken E F A F i G A B ii H D CH CH CH CH CH C FiedelCafts AC ae ab Vol. No.
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19 A X A A A A A A B A B A A A A A A Vol. No.
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21 A B B A B C A B C γ B γ α α β α β C Vol. No.
22 B A A A B B A B A A B C Vol. No.
23 A A A A B B A Vol. No.
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25 CBA CBA CBA CBA CBA CBA CBA CBA CBA CBA Vol. No.
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33 A B C C AC BC BCAC = a = + a = a a + a a a + a a AC k m = kλ a AC AC AC AC θ φ φ ψ θ ψ φ θ θ ψ θ ψ φ φ sinθ = nsinθ θ = θ + φ φ = φ θ = θ θ θ = nθ φ = φ = θ θ = n θ = n θ. 5 ( ) ( ) ( ) θ = FET wtv pq I = wtvqp VIR V be be b R = = = = I wtvqp wtqpμe wtqpμ V be R = = I wtv qp sat F = q( vi Bi ) = qvbi x z y Y Y Y E y Y F = q( E i vbi ) = y y y E = vb y V y I IB V = we = wvb = wb = y y wtqp tqp V y A C = ε t C Y C X C Y C X C Z C Vol. No.
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