When CH₄ is burnt in excess O₂ following products are formed,
CH₄ + 2 O₂ → CO₂ + 2 H₂O
According to equation 1 mole of CH₄ (16 g) reacts with 2 moles of O₂ to produce 1 mole of CO₂ and 2 moles of H₂O. Hence the products are,
1 mole of CO₂ and 2 moles of H₂O
Converting 1 mole CO₂ to grams;
As,
Mass = Moles × M.mass
Mass = 1 mol × 44 g.mol⁻¹
Mass = 40 g of CO₂
Converting 2 moles of H₂O to grams,
Mass = 2 mol × 18 g.mol⁻¹
Mass = 36 g of H₂O
Total grams of products;
Mass of CO₂ = 44 g
+ Mass of H₂O = 36 g
-------------
Total = 80 g of Product
Result:
80 grams of product is formed when 16 grams of CH₄ is burnt in excess of Oxygen.
Answer:
1425 mmHg.
Explanation:
The following data were obtained from the question:
Initial volume (V1) = 1.5 L
Initial pressure (P1) = 1 atm
Final volume (V2) = 0.8 L
Final pressure (P2) =?
Next, we shall determine the final pressure of the gas by using the Boyle's law equation as follow:
P1V1 = P2V2
1 × 1.5 = P2 × 0.8
1.5 = P2 × 0.8
Divide both side by 0.8
P2 = 1.5/0.8
P2 = 1.875 atm
Finally, we shall convert 1.875 atm to mmHg.
This can be obtained as follow:
1 atm = 760 mmHg
Therefore,
1.875 atm = 1.875 × 760 = 1425 mmHg.
Therefore, the new pressure of the gas is 1425 mmHg.
Answer:
Explanation:
Hello there!
In this case, when considering weak acids which have an associated percent dissociation, we first need to set up the ionization reaction and the equilibrium expression:
![HA\rightleftharpoons H^++A^-\\\\Ka=\frac{[H^+][A^-]}{[HA]}](https://tex.z-dn.net/?f=HA%5Crightleftharpoons%20H%5E%2B%2BA%5E-%5C%5C%5C%5CKa%3D%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
Now, by introducing x as the reaction extent which also represents the concentration of both H+ and A-, we have:
![Ka=\frac{x^2}{[HA]_0-x} =10^{-4.74}=1.82x10^{-5}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7Bx%5E2%7D%7B%5BHA%5D_0-x%7D%20%3D10%5E%7B-4.74%7D%3D1.82x10%5E%7B-5%7D)
Thus, it is possible to find x given the pH as shown below:
So that we can calculate the initial concentration of the acid:
![\frac{(1.82x10^{-5})^2}{[HA]_0-1.82x10^{-5}} =1.82x10^{-5}\\\\\frac{1.82x10^{-5}}{[HA]_0-1.82x10^{-5}} =1\\\\](https://tex.z-dn.net/?f=%5Cfrac%7B%281.82x10%5E%7B-5%7D%29%5E2%7D%7B%5BHA%5D_0-1.82x10%5E%7B-5%7D%7D%20%3D1.82x10%5E%7B-5%7D%5C%5C%5C%5C%5Cfrac%7B1.82x10%5E%7B-5%7D%7D%7B%5BHA%5D_0-1.82x10%5E%7B-5%7D%7D%20%3D1%5C%5C%5C%5C)
![[HA]_0=3.64x10^{-5}M](https://tex.z-dn.net/?f=%5BHA%5D_0%3D3.64x10%5E%7B-5%7DM)
Therefore, the percent dissociation turns out to be:
![\% diss=\frac{x}{[HA]_0}*100\% \\\\\% diss=\frac{1.82x10^{-5}M}{3.64x10^{-5}M}*100\% \\\\\% diss = 50\%](https://tex.z-dn.net/?f=%5C%25%20diss%3D%5Cfrac%7Bx%7D%7B%5BHA%5D_0%7D%2A100%5C%25%20%5C%5C%5C%5C%5C%25%20diss%3D%5Cfrac%7B1.82x10%5E%7B-5%7DM%7D%7B3.64x10%5E%7B-5%7DM%7D%2A100%5C%25%20%5C%5C%5C%5C%5C%25%20diss%20%3D%2050%5C%25)
Best regards!
The balanced chemical
reaction will be:
2H2O = 2H2 + O2
<span>We are given the amount of water used in the decomposition reaction. This will be our
starting point.</span>
<span>17.0 g H2O</span> (1 mol H2O/ 18.02 g H2O) (1 mol O2/2
mol <span>H2O</span>) ( 32.00 g O2/1mol O2) = 15.09 g O2
Percent yield = actual yield / theoretical yield x 100
<span>Percent yield =10.2 g / 15.09 g
x 100</span>
Percent yield = 67.58%
The first compound C6H12 is cyclohexane and the other compound C6H6 is benzene. They are both aromatic compounds. Cyclohexane does not have double bonds in its ring while benzene has three double bonds in its ring. This is why the formula for cyclohexane contains 12 carbon atoms while benzene only has 6.
