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>
When the substance is creating gases. Sometimes when it’s bubbling up
Answer:
For the first oxide, 1 g gives 0.888 g of copper.
Dividing by 0.888 tells us that 1.126 g gives 1 g of copper so has 0.126 g of oxygen.
For the second oxide, 1 g gives 0.798 g of copper.
Dividing by 0.798 tells us that 1.253 g gives 1 g of copper so has 0.253 g of oxygen.
So 1 g of copper combines with either 0.126 g or 0.253 g of oxygen.
Within the limits of experimental error, 0.253 is twice 0.126, confirming the law of multiple proportion.
Answer:
it should be momentum if im not mistaken
To calculate the moles of AgNO3 in a solution, we need to know the volume and concentration of the solution.
Moles of AgNO3 = Volume of AgNO3 solution (L) * concentration of AgNO3 solution (M or mole/L) = 1.50 L * 0.050 M = 0.075 mole.
So 0.075 moles of AgNO3 are present in 1.50 L of a 0.050 M solution.
