The reaction for the combustion of methane can be expressed as follows.
CH4 + 2O2 --> CO2 + 2H2O
We solve first for the amount of carbon dioxide in moles by dividing the given volume by 22.4L which is the volume of 1 mole of gas at STP.
moles of CO2 = (5.6 L) / (22.4 L/1 mole)
moles of CO2 = 0.25 moles
Then, we can see that every mole of carbon dioxide will need 1 mole of methane
moles methane = (0.25 moles CO2) x (1 moles O2/1 mole CO2)
= 0.25 moles CH4
Then, multiply this by the molar mass of methane which is 16 g/mole. Thus, the answer is 4 grams methane.
Starting in 1908, while a professor at the University of Chicago, Millikan worked on an oil-drop experiment in which he measured the charge on a single electron. J. J. Thomson had already discovered the charge-to-mass ratio of the electron.
Answer:
B. Hydrogen and oxygen molecules
General Formulas and Concepts:
<u>Chemistry - Reactions</u>
- Reactants are always on the left side of the arrow
- Products are always on the right side of the arrow
Explanation:
<u>Step 1: Define</u>
Reaction RxN: 2H₂ + O₂ → 2H₂O
<u>Step 2: Identify</u>
Reactants: H₂ and O₂
Products: H₂O
Answer:
Equilibrium concentration of 
is 12.5 M
Explanation:
Given reaction: 
Here, ![K_{c}=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}][H_{2}O]}](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5BC_%7B2%7DH_%7B5%7DOH%5D%7D%7B%5BC_%7B2%7DH_%7B4%7D%5D%5BH_%7B2%7DO%5D%7D)
where 
represents equilibrium constant in terms of concentration and species inside third bracket represent equilibrium concentrations
Here, ![[C_{2}H_{4}]=0.015M](https://tex.z-dn.net/?f=%5BC_%7B2%7DH_%7B4%7D%5D%3D0.015M)
, ![[C_{2}H_{5}OH]=1.69M](https://tex.z-dn.net/?f=%5BC_%7B2%7DH_%7B5%7DOH%5D%3D1.69M)
and 
So, ![[H_{2}O]=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}]\times K_{c}}=\frac{1.69}{0.015\times 9.0}=12.5M](https://tex.z-dn.net/?f=%5BH_%7B2%7DO%5D%3D%5Cfrac%7B%5BC_%7B2%7DH_%7B5%7DOH%5D%7D%7B%5BC_%7B2%7DH_%7B4%7D%5D%5Ctimes%20K_%7Bc%7D%7D%3D%5Cfrac%7B1.69%7D%7B0.015%5Ctimes%209.0%7D%3D12.5M)
Hence equilibrium concentration of is 12.5 M
