Natural gas is primarily composed of methane (CH4)
Natural gas is a naturally occurring hydrocarbon mixture which is primarily composed of Methane(CH4), but it also contains ethane,propane and heavier hydrocarbon. In addition it contain small amount of nitrogen, carbon dioxide,hydrogen sulfide and traces amount of water.
Formula:
f = C/λ
Where,
λ (Lambda) = Wavelength in meters
c = Speed of Light (299,792,458 m/s)
f = Frequency (MHz)
answer= 693964023.148148 MHz
Answer:
Kc = 6x10⁻⁶
Explanation:
For the reaction:
4NH₃(g) + 3O₂(g) ⇄ 2N₂(g) + 6H₂O(g)
Kc is defined as:
Kc =[N₂]² [H₂O]⁶ / [NH₃]⁴ [O₂]³
The equilibrium concentrations of the gases is -Because volume of the container is 1.00L-:
[N₂] = 2X = 1.96x10⁻³; <em>X = 9.8x10⁻⁴</em>
[H₂O] = 6X; 6ₓ9.8x10⁻⁴ = 5.88x10⁻³
[NH₃] = 0.0150M - 4X = 0.01108M
[O₂] = 0.0150M - 3X = 0.01206M
Replacing in Kc expression:
Kc =[1.96x10⁻³]² [5.88x10⁻³]⁶ / [0.01108M]⁴ [0.01206M]³
<h3>Kc = 6x10⁻⁶</h3>
There are three ways that scientists have proved that these sub-atomic particles exist. They are direct observation, indirect observation or inferred presence and predictions from theory or conjecture. Scientists in the 1800's were able to infer a lot about the sub-atomic world from chemistry.
Answer:
8.1 × 10² g
Explanation:
Step 1: Write the balanced equation
2 C₅₇H₁₁₀O₆ + 163 O₂ ⇒ 114 CO₂ + 110 H₂O
Step 2: Convert 1.6 lb of C₅₇H₁₁₀O₆ to g
We will use the conversion factor 1 lb = 453.592 g.
1.6 lb × 453.592 g/1 lb = 7.3 × 10² g
Step 3: Calculate the moles corresponding to 7.3 × 10² g of C₅₇H₁₁₀O₆
The molar mass of C₅₇H₁₁₀O₆ is 890.83 g/mol.
7.3 × 10² g × 1 mol/890.83 g = 0.82 mol
Step 4: Calculate the moles of water produced from 0.82 moles of C₅₇H₁₁₀O₆
The molar ratio of C₅₇H₁₁₀O₆ to H₂O is 2:110. The moles of H₂O produced are 110/2 × 0.82 mol = 45 mol
Step 5: Calculate the mass corresponding to 45 moles of H₂O
The molar mass of H₂O is 18.02 g/mol.
45 mol × 18.02 g/mol = 8.1 × 10² g
