Answer:
Sodium chloride solution:
First you need to calculate the mass of salt needed (done in the explanation), which is 58.44g. Then it have to be weighted in an analytical balance in a weighting boat and then transferred into a 2L volumetric flask that is going to be filled until the mark with distilled water.
Sulfuric acid dilution:
First you need to calculate the volume needed (done in the explanation), it is 16.6 mL. Using a graduated pipette one measures this volume and transfer it into a 2L volumetric flask that is already half filled with distilled water, and then one fills it until its mark.
Explanation:
Sodium chloride solution:
Each liter of a 0.500M solution has half mol, so 2L of said solution has 1 mol of salt. Sodium chloride molar mass is 58.44g/mol, so in 2L of solution there is 58.44g of salt. That`s the mass that`s going to be weighted and transferred to a 2L volumetric flask.
Sulfuric acid dilution:
This is the equation for dilution of solutions:
Where "c1" stands for the initial concentration (stock solution concentration), "v1" for the initial volume (volume of stock solution used), "c2" for the desired concentration and "v2" for the desired volume.
When we are diluting from a stock solution we want to know how much do we have to pipette from the stock solution into our volumetric flask. We do so by isolating the "v1" term from the dilution equation:
in this case that would be:
Answer:
The mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882
Explanation:
We are given that
Aqueous solution that contains 22.9% NaOH by mass means
22.9 g NaOH in 100 g solution.
Mass of NaOH(WB)=22.9 g
Mass of water =100-22.9=77.1
Na=23
O=16
H=1.01
Molar mass of NaOH(MB)=23+16+1.01=40.01
Number of moles =
Using the formula
Number of moles of NaOH

Molar mass of water=16+2(1.01)=18.02g
Number of moles of water

Now, mole fraction of NaOH
=

=0.882
Hence, the mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882
Answer:
Acetic acid 0,055M and acetate 0,095M.
Explanation:
It is possible to prepare a 0,15M buffer of acetic acid/acetate at pH 5,0 using Henderson-Hasselblach formula, thus:
pH = pka + log₁₀ [A⁻]/[HA] <em>-Where A⁻ is acetate ion and HA is acetic acid-</em>
Replacing:
5,0 = 4,76 + log₁₀ [A⁻]/[HA]
<em>1,7378 = [A⁻]/[HA] </em><em>(1)</em>
As concentration of buffer is 0,15M, it is possible to write:
<em>[A⁻] + [HA] = 0,15M </em><em>(2)</em>
Replacing (1) in (2):
1,7378[HA] + [HA] = 0,15M
2,7378[HA] = 0,15M
[HA] = 0,055M
Thus, [A⁻] = 0,095M
That means you need <em>acetic acid 0,055M</em> and <em>acetate 0,095M</em> to obtain the buffer you need.
i hope it helps!
The correct answer is oceanic crust, 80 km, Hope this helps let me know.
The designation "4" before Na2SO3 simply means that there are four molecules of sodium sulfite.
A molecule is formed when two or more atoms of the same or different elements come together. The formula of the molecule must reflect the number of atoms from which it is formed. The designation "4" before Na2SO3 simply means that there are four molecules of sodium sulfite.
The "The World on Turtle's Back" is a legend creation story. According to this story, the world was created on a giant turtle back. The folktale is common among the Iroquois Native Americans.
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