Izračunajte DeltaS ^ Theta na 25 ^ circC?

Odgovor:

D. # 17J #` # mol ^ -1 #

Pojasnilo:

Enačba za Gibbsovo brezplačno energijo je podana z:

# DeltaG = DeltaH-TDeltaS #

V tem primeru # DeltaS = (DeltaH-DeltaG) / T #

# DeltaS = (164000-159000) / (25 + 273) = 5000/298 = 16,8 ~ 17J # # K ^ -1 # # mol ^ -1- = D #

Odgovor:

# "D) 17 J / (K mol)" #

Pojasnilo:

Uporabite to enačbo

# "ΔG" ^ @ = "ΔH" ^ @ - "TΔS" ^ @ #

Po prerazporeditvi

# "ΔS" ^ @ = ("ΔH" ^ @ - "ΔG" ^ @) / ("T") = "164 kJ / mol - 159 kJ / mol" / "298 K" "17 J / (K mol) "#

Če 2sin theta + 3cos theta = 2 dokaže, da 3sin theta - 2 cos theta = ± 3?

Glej spodaj. Glede na rarr2sinx + 3cosx = 2 rarr2sinx = 2-3cosx rarr (2sinx) ^ 2 = (2-3cosx) ^ 2 rarr4sin ^ 2x = 4-6cosx + 9cos ^ 2x rarrcancel (4) -4cos ^ 2x = odpoved (4) - 6cosx + 9cos ^ 2x rarr13cos ^ 2x-6cosx = 0 rarrcosx (13cosx-6) = 0 rarrcosx = 0,6 / 13 rarrx = 90 ° Zdaj, 3sinx-2cosx = 3sin90 ° -2cos90 ° = 3

Poišči vrednost theta, če, Cos (theta) / 1 - sin (theta) + cos (theta) / 1 + sin (theta) = 4?

Theta = pi / 3 ali 60 ^ @ OK. Dobili smo: costheta / (1-sintheta) + costheta / (1 + sintheta) = 4 Zdaj ignoriramo RHS. costheta / (1-sintheta) + costheta / (1 + sintheta) (costheta (1 + sintheta) + costheta (1-sintheta)) / ((1-sintheta) (1 + sintheta)) (costheta ((1-sintheta) ) + (1 + sinteta)) / (1-sin ^ 2theta) (costheta (1-sintheta + 1 + sintheta)) / (1-sin ^ 2theta) (2costheta) / (1-sin ^ 2theta) Pitagorejska identiteta, sin ^ 2teta + cos ^ 2tea = 1. Torej: cos ^ 2theta = 1-sin ^ 2theta Zdaj, ko vemo, da lahko, pišemo: (2costheta) / cos ^ 2theta 2 / costheta = 4 costheta / 2 = 1/4 costheta = 1/2 theta = cos ^ - 1 (1/2)

Pokažite, da (1 + cos theta + i * sin theta) ^ n + (1 + cos theta - i * sin theta) ^ n = 2 ^ (n + 1) * (cos theta / 2) ^ n * cos ( n * theta / 2)?

Glej spodaj. Naj bo 1 + costheta + isintheta = r (cosalpha + isinalpha), tukaj r = sqrt ((1 + costheta) ^ 2 + sin ^ 2theta) = sqrt (2 + 2costheta) = sqrt (2 + 4cos ^ 2 (theta / 2) ) -2) = 2cos (theta / 2) in tanalpha = sintheta / (1 + costheta) == (2sin (theta / 2) cos (theta / 2)) / (2cos ^ 2 (theta / 2)) = tan (theta / 2) ali alfa = theta / 2, nato 1 + costheta-isintheta = r (cos (-alpha) + isin (-alpha)) = r (cosalpha-isinalpha) in lahko napišemo (1 + costheta + isintheta) ^ n + (1 + costheta-isintheta) ^ n z uporabo DE MOivrejevega izreka kot r ^ n (cosnalpha + isinnalpha + cosnalpha-isinnalpha) = 2r ^ ncosnalpha = 2 *