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NeX [460]
4 years ago
9

After performing a dilution calculation, you determine you need 25.0 milliliters of an aqueous stock solution to make 100.0 mill

iliters of a new solution. How would this be prepared?
Chemistry
2 answers:
IRINA_888 [86]4 years ago
8 0
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.
ch4aika [34]4 years ago
8 0

Answer:

You add 75 ml of water to a volumetric flask, along with the 25 ml of stock solution, to prepare a 100 ml dilution

Explanation:

Dilution is the procedure used to prepare a less concentrated solution from a more concentrated solution. It consists of adding solvent to an existing solution. Then the amount of solute does not vary, but the volume of the solvent does: when more solvent is added, the concentration of the solute decreases, as the volume of the solution increases.

In the case of a dilution, a Dilution Factor can be determined, which is a number that indicates how many times a solution must be diluted to obtain a lower concentration, such as:

Dilution factor = \frac{Vfinal}{Vinitial}

where Vfinal is the volume of the stock solution and Vinitial is the volume of the diluted solution.

In this case you need 25.0 ml of a stock solution to make 100.0 ml of a new diluted solution. So the dilution factor is:

Dilution factor=\frac{100}{25}

<em>Dilution factor=4</em>

This means that the initial solution must be diluted 4 times. Another way to see this is that for every 4 parts of this new diluted solution, 1 part is represented by the stock solution and 3 parts by the diluent water. Then, to prepare this solution, you must add 25.0 ml of stock solution to a volumetric flask, then add water until the total volume is equal to 100.0 ml.

So, you can do the calculation:

<em>100 ml - 25 ml = 75 ml </em>

This means that <u><em>you add 75 ml of water to a volumetric flask, along with the 25 ml of stock solution, to prepare a 100 ml dilution.  You diluted the stock solution by a factor of  4.</em></u>

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34.2 gram of sucrose is dissolved in 180 gram water. Calculate the number of hydrogen and oxygen atoms present in the solution.
Naddika [18.5K]

Explanation:

34.2g of C12H22O11 is dissolved in 180g of H20.

Molar mass of sucrose = 342g/mol

Moles of sucrose = 342 / 34.2 = 10 mol.

Molar mass of water = 18g/mol

Moles of water = 180 / 18 = 10 mol.

For hydrogen atoms, there are 22 * 10 in sucrose and 2 * 10 in water, which gives a total of 240.

For oxygen atoms, there are 11 * 10 in sucrose and 1 * 10 in water, which gives a total of 120.

8 0
3 years ago
Need help ASAP!!!! what is the value (angle) for the C=C=O bond in Ketene i.e. CH2=C=O
stepan [7]

Answer:

180^\circ by the VSEPR theory.

Explanation:

This question is asking for the bond angle of the \rm C=C=O bond in \rm H_2C=C=O. The VSEPR (valence shell electron pair repulsion) theory could help. Start by considering: how many electron domains are there on the carbon atom between these two bond?

Note that "electron domains" refer to covalent bonds and lone pairs collectively.

  • Each nonbonding pair (lone pair) of valence electrons counts as one electron domain.
  • Each covalent bond (single bond, double bond, or triple bond) counts as exactly one electron domain.

For example, in \rm H_2C=C=O, the carbon atom at the center of that \rm C=C=O bond has two electron domains:

  • This carbon atom has two double bonds: one \rm C=C bond and one \rm C=O bond. Even though these are both double bonds, in VSEPR theory, each of them count only as one electron domain.
  • Keep in mind that there are only four valence electrons in each carbon atom. It can be shown that all four valence electrons of this carbon atom are involved in bonding (two in each of the two double bonds.) Hence, there would be no nonbonding pair around this atom.

In VSEPR theory, electron domains around an atom repel each other. As a result, they would spread out (in three dimensions) as far away from each other as possible. When there are only two electron domains around an atom, the two electron domains would form a straight line- with one domain on each side of the central atom. (To visualize, consider the three atoms in this \rm C=C=O bond as three spheres on a stick. The central \rm C atom would be between the other \rm C atom and the \rm O atom.)

This linear geometry corresponds to a bond angle of 180^\circ.

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Answer:

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Explanation:

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