Answer:
4.93g are extracted
Explanation:
Partition coefficient (P) is defined as the ratio of solute dissolved in the organic solvent and the solute dissolved in the aqueous phase.
That is:
P = 7.5 = Concentration in dichloromethane / Concentration in water.
Knowing this, in the first extraction with 25mL of dichloromethane you will extract:
7.5 = (X/25mL) / (5g - X) / 100mL
Where X is the amount of compound A that is extracted.
7.5 = 100X / (125 - 25X)
937.5 - 187.5X = 100X
937.5 = 287.5X
3.26g of A are extracted in the first extraction.
In water will remain 5g - 3.26g = 1.74g
In the second extraction you will extract:
7.5 = (X/25mL) / (1.74g - X) / 100mL
7.5 = 100X / (43.5 - 25X)
326.25 - 187.5X = 100X
326.25 = 287.5X
1.13g are extracted in the second extraction.
And remain: 1.74g - 1.13g = 0.61g
In the third extraction you will extract:
7.5 = (X/25mL) / (0.61g - X) / 100mL
7.5 = 100X / (15.25 - 25X)
114.375 - 187.5X = 100X
114.375 = 287.5X
0.40g are extracted in the third extraction.
And remain: 0.61g - 0.40g = 0.21g
In the second extraction you will extract:
7.5 = (X/25mL) / (0.21g - X) / 100mL
7.5 = 100X / (5.25 - 25X)
39.375 - 187.5X = 100X
39.375 = 287.5X
0.14g are extracted in the fourth extraction.
Thus, after the three extractions you will extract: 0.14g + 0.40g + 1.13g + 3.26g = 4.93g are extracted
The process involves using the partitioncoefficient to determine how much of Compound A will prefer the dichloromethane solvent over the water. Following a calculation process through four rounds of extraction, it is concluded that approximately 4.999g of Compound A will be extracted using four 25mL portions of dichloromethane.
The partition coefficient of a compound is a measure of how much it prefers one solvent over another. Given that the partition coefficient of Compound A is 7.5 in dichloromethane with respect to water, we can predict how much of this compound could be extracted using four separate 25 mL portions of dichloromethane.
Here's the step-by-step calculation process:
In total, around 4.999g of compound A will be extracted using four 25mL portions of dichloromethane.
#SPJ3
Answer:
-1 Coulomb meter = -2.997 × 10²⁹ Debye
Explanation:
Given:
Coulomb meter = -1 CM
Find:
In debye
Computation:
We know that,
1 Coulomb meter = 299,792,458,178,090,000,000,000,000,000 Debye
So,
-1 Coulomb meter = -299,792,458,178,090,000,000,000,000,000 Debye
-1 Coulomb meter = -2.997 × 10²⁹ Debye
Answer:
the initial temperature of the iron sample is Ti = 90,36 °C
Explanation:
Assuming the calorimeter has no heat loss to the surroundings:
Q w + Q iron = 0
Also when the T stops changing means an equilibrium has been reached and therefore, in that moment, the temperature of the water is the same that the iron ( final temperature of water= final temperature of iron = T )
Assuming Q= m*c*( T- Tir)
mc*cc*(T-Tc)+mir*cir*(T - Tir) = 0
Tir = 20.3 °C + 300 g * 4.186 J/g°C * (20.3 C - 19 °C) / ( 51.9 g * 0.449 J/g°C )
Tir = 90.36 °C
Note :
- The specific heat capacity of water is assumed 1 cal/g°C = 4.186 J/g°C
- We assume no reaction between iron and water
To calculate the initial temperature of the iron sample, use the equation q = m * c * T, where q is the heat absorbed or released, m is the mass of the substance, c is the specific heat capacity, and T is the change in temperature which is 90.36 °C
To calculate the initial temperature of the iron sample, we can use the equation:
q = m * c * T
Where q is the heat absorbed or released, m is the mass of the substance, c is the specific heat capacity, andT is the change in temperature. In this case, we know the mass of the iron sample, the specific heat capacity of iron, and the change in temperature of the water. By rearranging the equation, we can solve for the initial temperature of the iron sample.
Thus,
Tir = 20.3 °C + 300 g * 4.186 J/g°C * (20.3 C - 19 °C) / ( 51.9 g * 0.449 J/g°C )
Tir = 90.36 °C
#SPJ12
Answer:
FCC.
Explanation:
Hello,
In this case, since the density is defined as:
Whereas n accounts for the number of atoms per units cell (2 for BCC and 4 for FCC), M the atomic mass of the element, Vc the volume of the cell and NA the Avogadro's number. Thus, for both BCC and FCC, the volume of the cell is:
Hence, we compute the density for each crystal structure:
Therefore, since the density computed as a FCC crystal structure matches with the actual density, we conclude rhodium has a FCC crystal structure.
Regards.
Answer:
Explanation:
Hello,
In this case, considering the reaction, we can compute the Gibbs free energy of reaction at each temperature, taking into account that the Gibbs free energy for the diatomic element is 0 kJ/mol:
Thus, at 2000 K:
And at 3000 K:
Next, since the relationship between the equilibrium constant and the Gibbs free energy of reaction is:
Thus, at each temperature we obtain:
In such a way, we can also conclude that at 2000 K reaction is unfavorable (K<1) and at 3000 K reaction is favorable (K>1).
Best regards.
Answer:
0,12 μmol/L of MgF₂
Explanation:
Preparation of solutions is a common work in chemist's life.
In this porblem says that you measure 0,00598 μmol of MgF₂ in 50,0 mL of water and you must calculate concentration in μmol/L
You have 0,00598 μmol but not Liters.
To obtain liters you sholud convert mL to L, knowing 1000mL are 1 L, thus:
50,0 mL (1L/1000mL) = 0,05 L of water.
Thus, concentration in μmol/L is:
0,00598 μmol / 0,05 L = 0,12 μmol/L -The problem request answer with two significant digits-
I hope it helps!
The chemical equation presented in option A follows the law of conservation of mass.
The principle of conservation of mass states, mass can neither be created nor destroyed but can be transformed from one form to another.
A reaction that follows the law of conservation of mass, must have equal number of moles each elements in reactants side and products side.
Only option A follows the law of conservation of mass;
Thus, we can conclude that the chemical equation presented in option A follows the law of conservation of mass.
Learn more here:brainly.com/question/13383562
Answer:
Option A
Explanation:
The expression that obeys the law of conservation of mass is choice A;
2LiOH + H₂CO₃ → Li₂CO₃ + 2H₂O
According to the law of conservation of mass; "in a chemical reaction, matter is neither created nor destroyed". By this law, mass is usually conserved.
The equation shows that mass is conserved because the number of moles of each specie is found on both sides
Number of moles
Li O H C
Reactants 2 5 4 1
Products 2 5 4 1
This shows that mass is indeed conserved.