<u>Answer:</u> The molarity of solution is 1.08 M
<u>Explanation:</u>
We are given:
(m/m) of phenol = 1.40 %
This means that 1.40 g of phenol is present in 100 g of solution.
To calculate volume of solution, we use the equation:
![\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}](https://tex.z-dn.net/?f=%5Ctext%7BDensity%20of%20substance%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20substance%7D%7D%7B%5Ctext%7BVolume%20of%20substance%7D%7D)
Density of solution = 0.9956 g/mL
Mass of solution = 100 g
Putting values in above equation, we get:
![0.9956g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=100.442mL](https://tex.z-dn.net/?f=0.9956g%2FmL%3D%5Cfrac%7B100g%7D%7B%5Ctext%7BVolume%20of%20solution%7D%7D%5C%5C%5C%5C%5Ctext%7BVolume%20of%20solution%7D%3D100.442mL)
To calculate the molarity of solution, we use the equation:
![\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20the%20solution%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20solute%7D%5Ctimes%201000%7D%7B%5Ctext%7BMolar%20mass%20of%20solute%7D%5Ctimes%20%5Ctext%7BVolume%20of%20solution%20%28in%20mL%29%7D%7D)
We are given:
Mass of solute (phenol) = 1.40 g
Molar mass of phenol = 94.11 g/mol
Volume of solution = 100.442 mL
Putting values in above equation, we get:
![\text{Molarity of solution}=\frac{1.40g\times 1000}{94.11g/mol\times 100.442mL}\\\\\text{Molarity of solution}=0.15M](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20solution%7D%3D%5Cfrac%7B1.40g%5Ctimes%201000%7D%7B94.11g%2Fmol%5Ctimes%20100.442mL%7D%5C%5C%5C%5C%5Ctext%7BMolarity%20of%20solution%7D%3D0.15M)
Hence, the molarity of solution is 0.15 M
Answer:
the second choice
Explanation:
hi! im not 100% sure but i do think its particles close together and moving past each other ( the 2nd choice) sorry its wrong
Answer:
D.When the boiling points of the components are different
Explanation:
I believe the formal name for this procedure is actually distillation.
Answer:
Aluminum metal
Explanation:
In order to properly answer this or a similar question, we need to know some basic rules about galvanic cells and standard reduction potentials.
First of all, your strategy would be to find a trusted source or the table of standard reduction potentials. You would then need to find the half-equations for aluminum and gold reduction:
![Al^{3+}+3e^-\rightarrow Al; E^o=-1.66 V](https://tex.z-dn.net/?f=Al%5E%7B3%2B%7D%2B3e%5E-%5Crightarrow%20Al%3B%20E%5Eo%3D-1.66%20V)
![Au^{3+}+3e^-\rightarrow Au; E^o=1.50 V](https://tex.z-dn.net/?f=Au%5E%7B3%2B%7D%2B3e%5E-%5Crightarrow%20Au%3B%20E%5Eo%3D1.50%20V)
Since we have a galvanic cell, the overall reaction is spontaneous. A spontaneous reaction indicates that the overall cell potential should be positive.
Since one half-equation should be an oxidation reaction (oxidation is loss of electrons) and one should be a reduction reaction (reduction is gain of electrons), one of these should be reversed.
Thinking simply, if the overall cell potential would be obtained by adding the two potentials, in order to acquite a positive number in the sum of potentials, we may only reverse the half-equation of aluminum (this would change the sign of E to positive):![Al\rightarrow Al^{3+}+3e^-; E^o=1.66 V\\Au^{3+}+3e^-\rightarrow Au; E^o=1.50 V](https://tex.z-dn.net/?f=Al%5Crightarrow%20Al%5E%7B3%2B%7D%2B3e%5E-%3B%20E%5Eo%3D1.66%20V%5C%5CAu%5E%7B3%2B%7D%2B3e%5E-%5Crightarrow%20Au%3B%20E%5Eo%3D1.50%20V)
Notice that the overall cell potential upon summing is:
![E_{cell}=1.66 V + 1.50 V=3.16 V](https://tex.z-dn.net/?f=E_%7Bcell%7D%3D1.66%20V%20%2B%201.50%20V%3D3.16%20V)
Meaning we obey the law of galvanic cells.
Since oxidation is loss of electrons, notice that the loss of electrons takes place in the half-equation of aluminum: solid aluminum electrode loses 3 electrons to become aluminum cation.
Mass of oxygen : 54 g
<h3>Further explanation
</h3>
Conservation of mass applies to a closed system, where the masses before and after the reaction are the same
Reaction
![\tt C+O_2\rightarrow CO_2](https://tex.z-dn.net/?f=%5Ctt%20C%2BO_2%5Crightarrow%20CO_2)
mass before reaction = mass after reaction
mass C + mass O₂= mass CO₂
![\tt 18~g+mass~O_2=72~g\\\\mass~O_2=72-18=\boxed{\bold{54~g}}](https://tex.z-dn.net/?f=%5Ctt%2018~g%2Bmass~O_2%3D72~g%5C%5C%5C%5Cmass~O_2%3D72-18%3D%5Cboxed%7B%5Cbold%7B54~g%7D%7D)