D. apply downward force to the lily pad, pushing itself upwards.
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 :)
When The rate of effusion is inversely proportional to the √molar mass of the substance.
and we have R(He) = 1L / 4.5 min so,
R(He)/R(Cl2) = (molar mass of Cl2/ molar mass of He)^0.5
and when we have the molar mass of Cl2 = 70.9 & the molar mass of He = 4
so by substitution:
(1L/4.5 min)/ R(Cl2) = (70.9 / 4)^0.5
(1L/4.5 min) / R(Cl2) = 4.21
∴R(Cl2) = (1L/4.5 min) / 4.21 = 1L/ (4.5*4.21)min = 1 L / 18.945 min
∴Cl2 will take 18.945 min for 1 L to effuse under identical conditions
Answer:
(a) 1s2 2s1
Explanation:
Electron configurations of atoms are in their ground state when the electrons completely fill each orbital before starting to fill the next orbital.
<h3><u>
Understanding the notation</u></h3>
It's important to know how to read and interpret the notation.
For example, the first part of option (a) says "1s2"
- The "1" means the first level or shell
- The "s" means in an s-orbital
- The "2" means there are 2 electrons in that orbital
<h3><u>
</u></h3><h3><u>
Other things to know about electron orbitals</u></h3>
It important to know which orbitals are in each shell:
- In level 1, there is only an s-orbital
- In level 2, there is an s-orbital and a p-orbital
- in level 3, there is an s-orbital, a p-orbital, and a d-orbital <em>(things get a little tricky when the d-orbitals get involved, but this problem is checking on the basic concept -- not the higher level trickery)</em>
So, it's also important to know how many electrons can be in each orbital in order to know if they are full or not. The electrons should fill up these orbitals for each level, in this order:
- s-orbitals can hold 2
- p-orbitals can hold 6
- d-orbitals can hold 10 <em>(but again, that's beyond the scope of this problem)</em>
<h3><u>
Examining how the electrons are filling the orbitals</u></h3>
<u>For option (a):</u>
- the 1s orbital is filled with 2, and
- the 2s orbital has a single electron in it with no other orbitals involved.
This is in it's ground state.
<u>For option (b):</u>
- the 1s orbital is filled with 2,
- the 2s orbital is filled with 2,
- the 2p orbital has 5 (short of a full 6), and
- the 3s orbital has a single electron in it.
Because the 3s orbital has an electron, but the lower 2p before it isn't full. This is NOT in it's ground state.
<u>For option (c):</u>
- the 1s orbital is filled with 2,
- the 2s orbital has 1 (short of a full 2), and
- the 2p orbital is filled with 6
Although the 2p orbital is full, since the 2s orbital before it was not yet full, this is NOT in it's ground state.
<u>For option (d):</u>
- the 1s orbital has 1 (short of a full 2), and
- the 2s orbital is filled with 2
Again, despite that the final orbital (in this case, the 2s orbital), is full, since the 1s orbital before it was not yet full, this is NOT in it's ground state.
To find out if a dye is a mixture or not, you can make use of the chromatography method of separation.
If the dye is a mixture, the different elements inside the mixture will be separated out during chromatography and they will form different bands on the cinematographic paper but if the dye is composed of only one element, only one band will be formed on the cinematographic paper.