Answer:
0.774g of ethanol
0.970mL of ethanol
Explanation:
Molality is an unit of concentration defined as the ratio between moles of solute and kg of solvent.
In the problem, you need to prepare a 1.2m solution of ethanol (Solute) in t-butanol (solvent).
14.0g of butanol are <em>0.014kg </em>and as you want to prepare the 1.2m solution, you need to add:
0.014kg × (1.2moles / kg) = 0.0168 moles of solute = Moles of ethanol
To convert moles of ethanol to mass you require molar mass (Molar mass ethanol, C₂H₅OH = 46.07g/mol). Thus, mass of 0.0168 moles are:
0.0168moles Ethanol ₓ (46.07g / mol) =
<h3>0.774g of ethanol</h3>
And to convert mass in g to mL you require density of the substance (Density of ethanol = 0.798g/mL):
0.774g ₓ (1mL / 0.798g) =
<h3>0.970mL of ehtanol</h3>
I don't know how well known/accepted this is (it's in my textbook so I'm guessing it's right), but Sulphur has two forms - the alpha and beta forms ,apparently gamma sulphur exists as well.
The alpha form is rhombic, yellow in color and has a MP of 385.8 K. The beta form is colorless and has a MP of 393 K and is formed by melting rhombic sulphur and cooling it till a crust forms on top. Poke a hole and pour out the liquid inside and you get beta sulphur. The transition point is 369K - below it, alpha sulphur is stable and above it, beta sulphur is stable. Both have helped. I had to pull out an old textbook and that's something that I don't usually do.
Answer : The concentration of
and
are
and
respectively.
Solution : Given,
pH = 4.10
pH : pH is defined as the negative logarithm of hydronium ion concentration.
Formula used : ![pH=-log[H_3O^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH_3O%5E%2B%5D)
First we have to calculate the hydronium ion concentration by using pH formula.
![4.10=-log[H_3O^+]](https://tex.z-dn.net/?f=4.10%3D-log%5BH_3O%5E%2B%5D)
![[H_3O^+]=antilog(-4.10)](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3Dantilog%28-4.10%29)
![[H_3O^+]=7.94\times 10^{-5}](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D7.94%5Ctimes%2010%5E%7B-5%7D)
Now we have to calculate the pOH.
As we know, 


Now we have to calculate the hydroxide ion concentration.
![pOH=-log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D)
![9.9=-log[OH^-]](https://tex.z-dn.net/?f=9.9%3D-log%5BOH%5E-%5D)
![[OH^-]=antilog(-9.9)](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3Dantilog%28-9.9%29)
![[OH^-]=1.258\times 10^{-10}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.258%5Ctimes%2010%5E%7B-10%7D)
Therefore, the concentration of
and
are
and
respectively.
4. Isotopes of the same element
When the atomic number increases The atomic size will be larger.