A is obviously out because it leads to a volume of 125.0 milliliters of the new solution and gives you a lower concentration than you were aiming for.
D is out because you are adding 75 milliliters of the stock solution, so your concentration would be too high. You only need 25.0 milometers of stock solution per 100 milliliters of the new solution.
C is also out because it leads to 50.0 milliliters stock solution per 100 milliliters of the new solution and hence the wrong concentration.
B is by default the correct answer. It also details the correct technique. First you add the stock solution (This you know from your calculations to be 25 milliliters.) then you add the water up to the volume you needed. (Because the calculations only tell you the total volume of water not what you need to add) You also add the water last so you can rinse the neck of the flask to make sure you also get all the stock solution residue into the stock solution.
I would add the final step of stirring, but B is the only answer that can be correct.
Answer:
This is all true if the atom has to be neutral.
Also what does V mean?
Helium: one shell with 2 neutrons and 2 protons in the center, with 2 electrons in the first shell.
Lithium: two shells with 4 neutrons and 3 protons in the center, with 2 electrons in the first shell, and 1 electron in the second shell.
Nitrogen: two shells with 7 neutrons and 7 protons in the center, with 2 electrons in the first shell, and 5 electrons in the second shell.
Flourine: two shells with 9 protons and 10 neutrons in the center, with 2 electrons in the first shell, and 7 electrons in the second shell.
Neon: two shells with 10 neutrons and 10 protons in the center, with 2 electrons in the first shell, and 8 electrons in the second shell.
Boron: two shells with 6 neutrons and 5 protons in the center, with 2 electrons in the first shell, and 3 electrons in the second shell.
Answer:
Explanation:
A sigma bond is formed when a hybrid orbital sp overlaps with another hybrid orbital or with s- or p- orbital
Ethylene has a structural formula of H - C≡ C- H
At the ground state; we have :
C | ⇅ | |↑ | ↑
2s 2p
At the excited state, we have:
C | ↑ | |↑ | ↑ | ↑
the hybrid orbital
C = | ↑ | ↑ |
2sp
Answer:
1 mole represents 6.023×1023 particles.
1 mole of iodine atom= 6.023×1023
Given 127.0g of iodine.
no. of iodine atom = 1 mole of iodine
1mole of magnesium = 24g of Mg = 6.023×1023no.of Mg
Given 48g of Mg = 2×6.023×1023
no. of Mg = 2 moles of Mg
1 mole of chlorine atom= 6.023× 1023
no. of chlorine atom = 35.5g of chlorine atom
Given 71g of chlorine atom=2× 6.023× 1023
no. of chlorine atom = 6.023×1023
2 moles of chlorine atom.
Given that 4g of hydrogen atom.
will be equal to 4 × 6.023 × 1023
no. of atoms of hydrogen= 4 moles of hydrogen atom.
