Answer:
I can use a dichotomous key. It helps me classify objects by sorting it out with "yes" and "no" questions.
or
I can use a Punnett Square. It helps me classify what genes the offspring will receive simply by figuring out the recessive and dominant genes as well as the hetzygous and homzygous.
Now give an example of which ever chart you choose by drawing it if that is required. For the Punnett Square label each of the squares Top right Hetzygous, top left dominant, bottom left recessive, bot-tom right homzygous. And for the dichotomous key put a 5-7 length branch showing the animals that have fur, can breathe under water, what cannot or doesn't have those traits. or something similar
Hopefully this helps :)
Explanation:
Apply the mass of balance as follows.
Rate of accumulation of water within the tank = rate of mass of water entering the tank - rate of mass of water releasing from the tank
[/tex]\frac{dh}{dt} + \frac{0.01}{0.01}h[/tex] = 
+ h = 1
= 1 - h
= dt
= t + C
Given at t = 0 and V = 0
= 0
or, h = 0
-ln(1 - h) = t + C
Initial condition is -ln(1) = 0 + C
C = 0
So, -ln(1 - h) = t
or, t = 
........... (1)
(a) Using equation (1) calculate time to fill the tank up to 0.6 meter from the bottom as follows.
t =
t = 
= 
= 0.916 seconds
(b) As maximum height of water level in the tank is achieved at steady state that is, t = 
.
1 - h = exp (-t)
1 - h = 0
h = 1
Hence, we can conclude that the tank cannot be filled up to 2 meters as maximum height achieved is 1 meter.
Answer:
- <u>Cadmium has larger atomic radius than sulfur.</u>
Explanation:
Down a period, atomic radii decrease from left to right due to the increase in the number of protons and electrons across a period: when a proton is added the pull of the electrons towards the nucleus is larger, so the size of the atom decreases.
Hence, you can compare the elements that belong to a same period and predict that the atom with lower atomic number (number of protons) will haver larger atomic radius. With that:
- Oxygen and fluorine are in the period 3, being oxygen to the left of fluorine, so oxygen is larger than fluorine.
- Sulfur and chlorine are in the period 4, being sulfur to the left of chlorine, so sulfur is larger than chlorine.
Now see whan happens down a group. Atomic radius increases from top to bottom within a group due to electron shielding. That permits you to compare the size of the elements in a group:
- Fluorine and chlorine are in the same group (17), with chlorine directly below fluorine, so the atomic radius of chlorine is larger than the atomic radius of fluorine.
- Sulfur and oxygen are in the same group (16), with sulfur directlly below oxygen, so sulfur the atomic radius of sulfur is larger than the atocmi radius of oxygen.
So far, you can rank the atomic radius of sulfur, chlorine, fluorine, and oxygen, in increasing order as:
- O < F < Cl < S, concluding that O, F, and Cl have smaller atomic radius than S.
Cadmiun, Cd, is to the left and below sulfur, so both electron shielding (down a group) and increase of the number of protons (down a period) lead to predict the cadmium has a larger atomic radius than sulfur.
So we end up with a total of four oxygen atoms for this calcium acetate unit and guys that truly it for this one.
Answer:
s = 4.41 g/L.
Explanation:
¡Hola!
En este caso, considerando el escenario dado, se hace necesario para nosotros saber que la posible reacción de disociación la experimenta el cloruro de plomo (II) como se muestra a continuación:
Lo cual hace que la expresión de equilibrio se calcule como:
![Ksp=[Pb^{2+}][Cl^-]^2](https://tex.z-dn.net/?f=Ksp%3D%5BPb%5E%7B2%2B%7D%5D%5BCl%5E-%5D%5E2)
Y que en términos de la solubilidad molar, s, se resuelve como:
![1.6x10^{-5}=s(2s)^2\\\\1.6x10^{-5}=4s^3\\\\s=\sqrt[3]{\frac{1.6x10^{-5}}{4} } \\\\s=0.0159molPbCl_2/L](https://tex.z-dn.net/?f=1.6x10%5E%7B-5%7D%3Ds%282s%29%5E2%5C%5C%5C%5C1.6x10%5E%7B-5%7D%3D4s%5E3%5C%5C%5C%5Cs%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B1.6x10%5E%7B-5%7D%7D%7B4%7D%20%7D%20%5C%5C%5C%5Cs%3D0.0159molPbCl_2%2FL)
Ahora, convertimos este valor a g/L al multiplicarlo por la masa molar del cloruro de plomo (II):
¡Saludos!
