Answer:
179 L of CO2
Explanation:
Given the equation of the reaction;
C2H6(g) + 7/2 O2(g) -------> 2CO2(g) + 3H2O(g)
Now 1 mole of ethane yields 2 moles of CO2 from the balanced reaction equation
1 mole of a gas occupies 22.4 L volume so,
22.4 L of ethane yields 44.8 L of CO2
89.5 L of ethane yields 89.5 * 44.8/22.4 = 179 L of CO2
Answer:
when lit, the atoms of different chemical compounds inside the firework absorb energy, when the electrons fall back to lower energy levels, they give off colored light. Depending upon the chemical used, different colors are produced. neon lights or signs are glass tubes filled with neon gas. When the light is plugged in, the electrical energy causes the electrons of neon to jump to higher energy levels, when they go to lower levels, colored light is given off. The neat thing is that the neon never gets used up. The only thing that gets used up is the electricity. The electrons don't get destroyed and can be used over and over again jumping levels. Neon glows with an orange light, but you know that neon signs can be lots of different colors. The variations come when different gases other than neon are used such as argon, or krypton or when the tube is painted so that with the light produced it produces a certain color.
Explanation:
Answer:

Explanation:
Hello,
In this case, for the given reaction, the equilibrium constant turns out:
![Keq=\frac{[B]}{[A]}=\frac{0.5M}{1.5M} =1/3](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BB%5D%7D%7B%5BA%5D%7D%3D%5Cfrac%7B0.5M%7D%7B1.5M%7D%20%3D1%2F3)
Nonetheless, we are asked for the reverse equilibrium constant that is:

Which is greater than one.
In such a way, the Gibbs free energy turns out:

Now, since the reverse equilibrium constant is greater than zero its natural logarithm is positive, therefore with the initial minus, the Gibbs free energy is less than zero, that is, negative.
Answer:
0 meters
Explanation:
I guessed on the test and I got it right lol.
First, let us find the corresponding amount of moles H₂ assuming ideal gas behavior.
PV = nRT
Solving for n,
n = PV/RT
n = (6.46 atm)(0.579 L)/(0.0821 L-atm/mol-K)(45 + 273 K)
n = 0.143 mol H₂
The stoichiometric calculations is as follows (MW for XeF₆ = 245.28 g/mol)
Mass XeF₆ = (0.143 mol H₂)(1 mol XeF₆/3 mol H₂)(245.28 g/mol) = <em>11.69 g</em>