Answer:
True
Explanation:
Well,I hope it helps....
Just correct me if I'm wrong..
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>
Answer:
b) The molecule has a molecular weight under 200 g/mole
Explanation:
The molecule has a molecular weight under 200 g/mole is the primary requirement for a molecule to be analyzed by Gas Chromatography.
Answer:
10425 J are required
Explanation:
assuming that the water is entirely at liquid state at the beginning , the amount required is
Q= m*c*(T final - T initial)
where
m= mass of water = 25 g
T final = final temperature of water = 100°C
T initial= initial temperature of water = 0°C
c= specific heat capacities of water = 1 cal /g°C= 4.186 J/g°C ( we assume that is constant during the entire temperature range)
Q= heat required
therefore
Q= m*c*(T final - T initial)= 25 g * 4.186 J/g°C * (100°C- 0°C) = 10425 J
thus 10425 J are required
