A. a wide variety of hunting techniques
Explanation:
To help the fisher population to survive in a changing environment, they have to develop a wide variety of hunting techniques.
The new fisher population have to develop a wide range of strategies to satisfy its carnivorous instincts.
- Most predators are carnivores with a wide range of hunting skills
- They must be fast, subtle, good sighted and well adapted body part to hunt of preys.
- As they transcends different environments, they must have good and diverse hunting skills to give them a competitive advantage.
Learn more:
Animal domestication brainly.com/question/3807452
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Answer:
15 electrones: 1S²2S²2P⁶3S²3P³. Fósforo
27 electrones: 1S²2S²2P⁶3S²3P⁶4S²3d⁷ - Cobalto.
56 electrones: 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s² - Bario
49 electrones: 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p¹ - Indio
Explanation:
Para llenar los orbitales electrónicos de los distintos átomos debemos hacer uso de la regla de llenado electrónico de Aufbau. Por ejemplo, para el átomo con 15 electrones, la configuración electrónica es:
1S²2S²2P⁶3S²3P³. 2+2+6+2+3 = 15 electrones
Si elemento es neutro, tiene 15 protones. Es decir, es el fósforo, P.
27 electrones:
1S²2S²2P⁶3S²3P⁶4S²3d⁷ - Cobalto.
56 electrones:
1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s² - Bario
49 electrones:
1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p¹ - Indio
Answer:
The difference in the absorbance of the sample and reference is too small at A = 0.05.3. If the absorbance is too high (A = 2.5),
Answer:
I think this answer should be 2.33 g H2O
Answer:
a) The value of the
is 0.52.
b)Concentration of all species when equilibrium reestablishes:
![[CO_2] = 0.4748 M](https://tex.z-dn.net/?f=%5BCO_2%5D%20%3D%200.4748%20M)
![[H_2] = 0.0198 M](https://tex.z-dn.net/?f=%5BH_2%5D%20%3D%200.0198%20M)
![[CO] = 0.0752 M](https://tex.z-dn.net/?f=%20%5BCO%5D%20%3D%200.0752%20M)
![[H_2O] =0.0652 M](https://tex.z-dn.net/?f=%5BH_2O%5D%20%3D0.0652%20M)
Explanation:
a) 
Equilibrium concentration of species :
![[CO] = 0.050 M, [H_2] = 0.045 M, [CO_2] = 0.086 M, and [H_2O] = 0.040 M](https://tex.z-dn.net/?f=%5BCO%5D%20%3D%200.050%20M%2C%20%5BH_2%5D%20%3D%200.045%20M%2C%20%5BCO_2%5D%20%3D%200.086%20M%2C%20and%20%5BH_2O%5D%20%3D%200.040%20M)
The expression of equilibrium constant is given as :
![\K_c=\frac{[CO][H_2O]}{[CO_2][H_2]}](https://tex.z-dn.net/?f=%5CK_c%3D%5Cfrac%7B%5BCO%5D%5BH_2O%5D%7D%7B%5BCO_2%5D%5BH_2%5D%7D)

b)

Initially:
0.50 M 0.045 M 0.050 M 0.040 M
At equilibrium ;
(0.50-x) M (0.045-x) M (0.050+x) M (0.040+x) M
![K_c=\frac{[CO][H_2O]}{[CO_2][H_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO%5D%5BH_2O%5D%7D%7B%5BCO_2%5D%5BH_2%5D%7D)

Solving fro x;
x = 0.0252
Concentration of all species when equilibrium reestablishes:
![[CO_2] = (0.50-x) M=(0.50-0.0252)M = 0.4748 M](https://tex.z-dn.net/?f=%5BCO_2%5D%20%3D%20%280.50-x%29%20M%3D%280.50-0.0252%29M%20%3D%200.4748%20M)
![[H_2] = (0.045-x) M= (0.045-0.0252) M=0.0198 M](https://tex.z-dn.net/?f=%5BH_2%5D%20%3D%20%280.045-x%29%20M%3D%20%280.045-0.0252%29%20M%3D0.0198%20M)
![[CO] = (0.050+x) M=(0.050+0.0252)M = 0.0752 M](https://tex.z-dn.net/?f=%20%5BCO%5D%20%3D%20%280.050%2Bx%29%20M%3D%280.050%2B0.0252%29M%20%3D%200.0752%20M)
![[H_2O] = (0.040+x) M=(0.040+0.0252) M=0.0652 M](https://tex.z-dn.net/?f=%5BH_2O%5D%20%3D%20%280.040%2Bx%29%20M%3D%280.040%2B0.0252%29%20M%3D0.0652%20M)