Answer:
Answer:
The mole ratio of C₄H₁₀ and CO₂ is 2 : 8, which simplifies to 1 : 4.
Explanation:
The mole ratio is the relative proportion of the moles of products or reactants that participate in the reaction according to the chemical equation.
The chemical equation given is:
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
Once you check that the equation is balanced, you can set the mole ratios for all the reactants and products. The coefficients used in front of each reactant and product, in the balanced chemical equation, tells the mole ratios.
In this case, they are: 2 mol C₄H₁₀ : 13 mol O₂ : 8 mol CO₂ : 10 mol H₂O
Since you are asked about the mole ratio of C₄H₁₀ and CO₂ it is:
2 mol C₄H₁₀ : 8 mol CO₂ , which dividing by 2, simplifies to
1 mol C₄H₁₀ : 4 mol CO₂, or
1 : 2.
Explanation:
The missing components in the table to the right are indicated with orange letters. Use the periodic table in the tools bar and this link Web Elements to fill in the corresponding values. A B C D E F G. 2. See answers. Log in to add ... F = 737.7kJ/mol. G = 495.8kJ/mol. Explanation: We are asked some of the ...
2 answers
I would say developed technologies which somehow were lost
like the stone age: the stone hammer, who used that afterwards in the roman empire era and stuff?
The Law of Conservation of Mass states that the mass of reactants entering a reaction must be equal to the mass of the products exiting it. In this case, we only have 2 reactants, Fe and S, and we only have 1 product, FeS. Therefore we expect the total mass of the Fe and S reactants to equal the mass of FeS. This gives us 112 g + 64 g = 176 g of FeS, which is choice D.
Answer:
V₂ = 15.3
Explanation:
Given data:
Initial volume = 12.0 L
Initial temperature = 20°C
Final temperature =100°C
Final volume = ?
Solution:
First of all we will convert the temperature into kelvin.
20°C + 273 = 293 K
100°C + 273 = 373 K
Formula:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 12.0 L × 373 K / 293 k
V₂ = 4476 L.K /293 k
V₂ = 15.3
V₂ = 1566 L.K / 298 K
V₂ = 5.3 L
