Answer:
The partial pressure of argon in the jar is 0.944 kilopascal.
Explanation:
Step 1: Data given
Volume of the jar of air = 25.0 L
Number of moles argon = 0.0104 moles
Temperature = 273 K
Step 2: Calculate the pressure of argon with the ideal gas law
p*V = nRT
p = (nRT)/V
⇒ with n = the number of moles of argon = 0.0104 moles
⇒ with R = the gas constant = 0.0821 L*atm/mol*K
⇒ with T = the temperature = 273 K
⇒ with V = the volume of the jar = 25.0 L
p = (0.0104 * 0.0821 * 273)/25.0
p = 0.00932 atm
1 atm =101.3 kPa
0.00932 atm = 101.3 * 0.00932 = 0.944 kPa
The partial pressure of argon in the jar is 0.944 kilopascal.
Answer: The concentration of KOH for the final solution is 0.275 M
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
where,
n = moles of solute
= volume of solution in ml = 150 ml
moles of solute =
Now put all the given values in the formula of molality, we get
According to the dilution law,
where,
= molarity of stock solution = 1.19 M
= volume of stock solution = 15.0 ml
= molarity of diluted solution = ?
= volume of diluted solution = 65.0 ml
Putting in the values we get:
Therefore, the concentration of KOH for the final solution is 0.275 M
Answer:
[EtOH] = 2.2M and Wt% EtOH = 10.1% (w/w)
Explanation:
1. Molarity = moles solute / Volume solution in Liters
=> moles solute = mass solute / formula weight of solute = 9.8g/46g·mol⁻¹ = 0.213mol EtOH
=> volume of solution (assuming density of final solution is 1.0g/ml) ...
volume solution = 9.81gEtOH + 87.5gH₂O = 97.31g solution x 1g/ml = 97.31ml = 0.09731 Liter solution
Concentration (Molarity) = moles/Liters = 0.213mol/0.09731L = 2.2M in EtOH
2. Weight Percent EtOH in solution (assuming density of final solution is 1.0g/ml)
From part 1 => [EtOH] = 2.2M in EtOH = 2.2moles EtOH/1.0L soln
= {(2.2mol)(46g/mol)]/1000g soln] x 100% = 10.1% (w/w) in EtOH.
Answer: C. ethanol
The enthalpy of combustion is the amount of heat produced when one mole of ethanol undergoes complete combustion at 25 ° C and 1 atmosphere pressure, yielding products also at 25 ° C and 1 atm.
<u>The enthalpy of combustion of the unknown compound is</u>
ΔH = - 320 kJ / 0.25 mol = - 1280 kJ / mol
<u>To choose a probable compound according to this combustion enthalpy, we must evaluate the deviation in relation to the values reported in the literature for the three probable compounds</u> (methane, ethylene and ethanol). The deviation (e%) will be calculated according to the following equation,
e% = ( | ΔHx - ΔH | / ΔHx ) x 100%
where ΔHx is the enthalpy of combustion of the probable compound.
The following table shows the combustion enthalpies of the probable compounds and their deviation in relation to the enthalpy of ΔH = - 1280 kJ / mol
Compound Enthalpy of combustion (kJ/mol) Deviation
Methane - 890.7 43.8%
Ehylene -1411.2 9.3%
Ethanol -1368.6 6.5%
According to the previous table, we can say that the most probable compound is ethanol, since it has the smallest deviation in relation to the experimental enthalpy value of combustion.
Answer:
See explanation
Explanation:
If the spot in TLC is below the solvent front, it will be observed that the spot, instead of being separated by the solvent as expected, will just dissolve away in the solvent and zero actual separation of the mixture is achieved.
If the solute is dissolved away instead of being separated by the solvent, then the experiment fails because no actual separation of the mixture is achieved.
Hence, in TLC, the spot must be applied above the solvent front so that the capillary movement of the solvent through the plate can lead to the eventual separation of the components of the mixture since the various components of the mixture will travel at different speeds through the plate.
Also, if the solvent is above the spot, the solvent may evaporate selectively from the points above the spot while separation is ongoing.
