What volume of carbon dioxide can be produced from the combustion of 2.0 L of methane at STP?
1 answer:
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The density of the wood : 1.45 g/cm³
<h3>Further explanation </h3>Density is a quantity derived from the mass and volume
Density is the ratio of mass per unit volume
With the same mass, the volume of objects that have a high density will be smaller than objects with a smaller type of density
The unit of density can be expressed in g/cm³ or kg/m³
Density formula:
ρ = density
m = mass
v = volume
- Volume of the wood
- density
I have attached the diagram related to this question.
Answer:
5A₂ + 5B .................> 4A₂B + A₂ + B
Explanation:
1- For the reactants:
We can note that:
i- we have 5 moles of substance A which is diatomic. This means that we have 5A₂
ii- we have 5 moles of substance B which is monoatomic. This means that we have 5B
Based on the above, for the reactants side, we have:
5A₂ + 5B
2- For the products:
We can note that:
i- we have 4 moles of a substance formed from the combination of two moles of substance A with one mole of substance B. This means that we have 4A₂B
ii- we have one mole of substance A that is diatomic. This means that we have A₂
iii- we have one mole of substance B that is monoatomic. This means that we have B
Based on the above, for the products side, we have:
4A₂B + A₂ + B
3- Combining reactants and products parts:
Combining the reactants side (from part 1) with the products side (from part 2), we can find that the initial equation is as follows:
5A₂ + 5B .................> 4A₂B + A₂ + B
4- Balancing the equation:
Taking a look at the initial equation we reached in part 3, we will find that:
Number of moles of substance A in reactants = 2*5 = 10 moles
Number of moles of substance A in products = 2*4 + 2*! = 10 moles
SUBSTANCE A IS BALANCED
Number of moles of substance B in reactants = 5*1 = 5 moles
Number of moles of substance B in products = 4*1 + 1 = 5 moles
SUBSTANCE B IS BALANCED
Based on the above, the final balanced equation is the same as the initial equation reached in part 3:
5A₂ + 5B .................> 4A₂B + A₂ + B
Hope this helps :)
Answer:
5A₂ + 5B .................> 4A₂B + A₂ + B
Explanation:
1- For the reactants:
We can note that:
i- we have 5 moles of substance A which is diatomic. This means that we have 5A₂
ii- we have 5 moles of substance B which is monoatomic. This means that we have 5B
Based on the above, for the reactants side, we have:
5A₂ + 5B
2- For the products:
We can note that:
i- we have 4 moles of a substance formed from the combination of two moles of substance A with one mole of substance B. This means that we have 4A₂B
ii- we have one mole of substance A that is diatomic. This means that we have A₂
iii- we have one mole of substance B that is monoatomic. This means that we have B
Based on the above, for the products side, we have:
4A₂B + A₂ + B
3- Combining reactants and products parts:
Combining the reactants side (from part 1) with the products side (from part 2), we can find that the initial equation is as follows:
5A₂ + 5B .................> 4A₂B + A₂ + B
4- Balancing the equation:
Taking a look at the initial equation we reached in part 3, we will find that:
Number of moles of substance A in reactants = 2*5 = 10 moles
Number of moles of substance A in products = 2*4 + 2*! = 10 moles
SUBSTANCE A IS BALANCED
Number of moles of substance B in reactants = 5*1 = 5 moles
Number of moles of substance B in products = 4*1 + 1 = 5 moles
SUBSTANCE B IS BALANCED
Based on the above, the final balanced equation is the same as the initial equation reached in part 3:
5A₂ + 5B .................> 4A₂B + A₂ + B
Hope this helps :)