Astronauts use "Astronomical Unit" to measure the distance between planets in our solar system "only". To measure long distances like between 2 galaxies, they use "light years". One Astronomical Unit = Average distance between sun and earth which is equal to 149.6 million kilometers. Ping me if you have any doubt.
Answer:
C. sorry if I'm wrong ...................
Answer:
Genotypes: Homozygous (GG)=50%, Heterozygous (Gg)=50%.
Phenotypes: Homozygous gray (GG)=50%, Heterozygous gray (Gg)=50% or just Gray=100%
Explanation:
Hello,
The Punnett square for this cross turns into:
![\left[\begin{array}{ccc}&G&g\\G&GG&Gg\\G&GG&Gg\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%26G%26g%5C%5CG%26GG%26Gg%5C%5CG%26GG%26Gg%5Cend%7Barray%7D%5Cright%5D)
It means that the genotypes and phenotypes are:
Genotypes: Homozygous (GG)=50%, Heterozygous (Gg)=50%.
Phenotypes: Homozygous gray (GG)=50%, Heterozygous gray (Gg)=50% or just Gray=100%
Best regards.
Answer:
K(48.5°C) = 1.017 E-8 s-1
Explanation:
- CH3Cl + H2O → CH3OH + HCl
at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1
at T2 = 48.5°C (321.5 K) ⇒ K2 = ?
Arrhenius eq:
- K(T) = A e∧(-Ea/RT)
- Ln K = Ln(A) - [(Ea/R)(1/T)]
∴ A: frecuency factor
∴ R = 8.314 E-3 KJ/K.mol
⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)
⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)
(1)/(2):
⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)
⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)
⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)
⇒ Ln (K1/K2) = - 3.422
⇒ K1/K2 = e∧(-3.422)
⇒ (3.32 E-10 s-1)/K2 = 0.0326
⇒ K2 = (3.32 E-10 s-1)/0.0326
⇒ K2 = 1.017 E-8 s-1
