Answer:
We need 0.375 mol of CH3OH to prepare the solution
Explanation:
For the problem they give us the following data:
Solution concentration 0,75 M
Mass of Solvent is 0,5Kg
knowing that the density of water is 1g / mL, we find the volume of water:
Now, find moles of 
are needed using the molarity equation:
therefore the solution is prepared using 0.5 L of H2O and 0.375 moles of CH3OH, resulting in a concentration of 0,75M
The correct answer for this question is this one: " a.The solution has a volume of 25 mL "
The observation that shows a quantitative observation is when you are talking about numeric data. Just like this one, <em>The solution has a volume of 25 mL </em>
Hope this helps answer your question and have a nice day ahead.
Answer:
I can't draw diagrams on this web site but I can do with numbers I think. So an electron is moved from n = 1 to n = 5. I'm assuming I've interpreted the problem correctly; if not you will need to make a correction. I'm assuming that you know the electron in the n = 1 state is the ground state so the 4th exited state moves it to the n = 5 level.
n = 5 4th excited state
n = 4 3rd excited state
n = 3 2nd excited state
n = 2 1st excited state
n = 1 ground state
Here are the possible spectral lines.
n = 5 to 4, n = 5 to 3, n = 5 to 2, n = 5 to 1 or 4 lines.
n = 4 to 3, 4 to 2, 4 to 1 = 3 lines
n = 3 to 2, 3 to 1 = 2 lines
n = 2 to 1 = 1 line. Add 'em up. I get 10.
b. The Lyman series is from whatever to n = 1. Count the above that end in n = 1.
c.The E for any level is -21.8E-19 Joules/n^2
To find the E for any transition (delta E) take E for upper n and subtract from the E for the lower n and that gives you delta E for the transition.
So for n = 5 to n = 1, use -Efor 5 -(-Efor 1) = + something which I'll leave for you. You could convert that to wavelength in meters with delta E = hc/wavelength. You might want to try it for the Balmer series (n ending in n = 2). I think the red line is about 650 nm.
Explanation:
Molecules with a plane of symmetry between the chiral centers are achiral and meso. From the given molecules (Picture attached) only (A) compound 1 is meso.
When compounds possess a plane of symmetry between the chiral centers they are called achiral or meso compounds. Among the given compounds (A) compound 1 have a plane of symmetry. So we can say compound one is a meso or achiral compound. Compounds two, three, and four have no plane of symmetry, as you can see in the structures attached. So all other compounds (compound 2, compound 3, and compound 4) except compound one are not meso or achiral.
You can also learn about meso compounds from the following question:
brainly.com/question/29022658
#SPJ4
Answer: Volatile
Explanation:
like ethanol(alcohol) it evaporates when you put it in the hand, really fast.
Some people think that it dries when it really evaporated at room temperature from liquid to gas
