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.
4p, 3d, 3p, 3s, 2p, 2s and 1s orbitals may be occupied during de-excitation.<span />
Answer: 
Explanation:
a) 
: This is a non polar covalent compound which are held by weak vanderwaal forces of attraction.
b) 
: This is a covalent compound which is polar due to the presence of lone pair of electrons and are held by dipole-dipole forces of attraction.
c) : These are joined by a special type of dipole dipole attraction called as hydrogen bond. It forms between electronegative nitrogen atom and hydrogen atom and is the strongest interaction.
d) 
: This is a covalent compound and is non polar which are held by weak vanderwaal forces of attraction.
e) 
: This is a covalent compound and is non polar which are held by weak vander waal forces of attraction.
Answer:
5250 grams or 5.25 kg of carbon monoxide and 375 grams of hydrogen are required to form 6 kg of methanol.
Explanation:
The balanced reaction:
CO (g) + 2 H₂ (g) -> CH₃OH (l)
By stoichiometry of the reaction, the following amounts of moles of each compound participate in the reaction:
- CO: 1 mole
- H₂: 2 moles
- CH₃OH: 1 mole
Being the molar mass of each compound:
- CO: 28 g/mole
- H₂: 1 g/mole
- CH₃OH: 32 g/mole
By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- CO: 1 mole* 28 g/mole= 28 grams
- H₂: 2 moles* 1 g/mole= 2 grams
- CH₃OH: 1 mole* 32 g/mole= 32 grams
Being 6 kg equivalent to 6000 grams (1 kg= 1000 grams), you can apply the following rules of three:
- If by stoichiometry 32 grams of methanol are formed from 28 grams of carbon monoxide, 6000 grams of methanol are formed from how much mass of carbon monoxide?
mass of carbon monoxide= 5250 grams= 5.25 kg
If by stoichiometry 32 grams of methanol are formed from 2 grams of hydrogen, 6000 grams of methanol are formed from how much mass of hydrogen?
mass of hydrogen= 375 grams
<u><em>5250 grams or 5.25 kg of carbon monoxide and 375 grams of hydrogen are required to form 6 kg of methanol. </em></u>
