Answer:
-5.51 kJ/mol
Explanation:
Step 1: Calculate the heat required to heat the water.
We use the following expression.
where,
- c: specific heat capacity
- m: mass
- ΔT: change in the temperature
The average density of water is 1 g/mL, so 75.0 mL ≅ 75.0 g.
Step 2: Calculate the heat released by the methane
According to the law of conservation of energy, the sum of the heat released by the combustion of methane (Qc) and the heat absorbed by the water (Qw) is zero
Qc + Qw = 0
Qc = -Qw = -22.0 kJ
Step 3: Calculate the molar heat of combustion of methane.
The molar mass of methane is 16.04 g/mol. We use this data to find the molar heat of combustion of methane, considering that 22.0 kJ are released by the combustion of 64.00 g of methane.
Answer:
Mole fraction of solute is 0.0462
Explanation:
To solve this we use the colligative property of lowering vapor pressure.
First of all, we search for vapor pressure of pure water at 25°C = 23.8 Torr
Now, we convert the Torr to mmHg. Ratio is 1:1, so 23.8 Torr is 23.8 mmHg.
Formula for lowering vapor pressure is:
ΔP = P° . Xm
Where ΔP = P' (Vapor pressure of solution) - P° (Vapor pressure of pure solvent)
Xm = mole fraction
24.9 mmHg - 23.8 mmHg = 23mmHg . Xm
Xm = (24.9 mmHg - 23.8 mmHg) / 23mmHg
Xm = 0.0462
Answer:
Lead is added to warm dilute nitric acid. When the carbonate has reacted with the warm acid, more carbonate is added until the carbonate is in excess.
Explanation:
DE = dH - PdV
<span>2 H2O(g) → 2 H2(g) + O2(g) </span>
<span>You can see that there are 2 moles of gas in the reactants and 3 moles of gas in the products. </span>
<span>1 moles of ideal gas occupies the same volume as 1 mole of any other ideal gas under the same conditions of temp and pressure. </span>
<span>Since it is done under constant temp and pressure that means the volume change will be equal to the volume of 1 mole of gas </span>
<span>2 moles reacts to form 3 moles </span>
<span>The gas equation is </span>
<span>PV = nRT </span>
<span>P = pressure </span>
<span>V = volume (unknown) </span>
<span>n = moles (1) </span>
<span>R = gas constant = 8.314 J K^-1 mol^-1 </span>
<span>- the gas constant is different for different units of temp and pressure (see wikki link) in this case temp and pressure are constant, and we want to put the result in an equation that has Joules in it, so we select 8.314 JK^-1mol^-1) </span>
<span>T = temp in Kelvin (kelvin = deg C + 273.15 </span>
<span>So T = 403.15 K </span>
<span>Now, you can see that PV is on one side of the equation, and we are looking to put PdV in our dE equation. So we can say </span>
<span>dE = dH -dnRT (because PV = nRT) </span>
<span>Also, since the gas constant is in the unit of Joules, we need to convert dH to Joules </span>
<span>dH = 483.6 kJ/mol = 483600 Joules/mol </span>
<span>dE = 483600 J/mol - (1.0 mol x 8.314 J mol^-1K-1 x 403.15 K) </span>
<span>dE = 483600 J/mol - 3351.77 J </span>
<span>dE = 480248.23 J/mol </span>
<span>dE = 480.2 kJ/mol </span>
