<u>Answer:</u> The entropy change of the liquid water is 63.4 J/K
<u>Explanation:</u>
To calculate the entropy change for same phase at different temperature, we use the equation:

where,
= Entropy change
= molar heat capacity of liquid water = 75.38 J/mol.K
n = number of moles of liquid water = 3 moles
= final temperature = ![95^oC=[95+273]K=368K](https://tex.z-dn.net/?f=95%5EoC%3D%5B95%2B273%5DK%3D368K)
= initial temperature = ![5^oC=[5+273]K=278K](https://tex.z-dn.net/?f=5%5EoC%3D%5B5%2B273%5DK%3D278K)
Putting values in above equation, we get:

Hence, the entropy change of the liquid water is 63.4 J/K
Answer: At STP, a mole of gas takes up 22.4 Liters. The 22.4 Liters/mole quantity can be derived from the Ideal Gas Law, PV = nRT, plugging in STP conditions for P and T, and solving for V/n, which gets 22.4 Liters/mole.
Explanation:
Answer:
A solution that is 0.10 M HCN and 0.10 M LiCN
Explanation:
- A good buffer system contains a weak acid and its salt or a weak base and its salt.
- In this case; A solution that is 0.10 M HCN and 0.10 M LiCN, would make a good buffer system.
- HCN is a weak acid, while LiCN is a salt of the weak acid, that is, CN- conjugate of the acid.
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:
on each side of the salt bridge, which is represented by a double vertical line
Explanation:
While writing a cell notation, the general convention is; anode || cathode. The anode and the cathode are separated by a double line. The anode is written on the lefthand side while the cathode is written on the righthand side.
The cell notation is a shorthand representation of a cell, hence any electrochemical cell can easily be produced based on its cell diagram.