Another product: CO₂
<h3>Further explanation</h3>
Given
Reaction
2C₄H₁₀ + 13O₂⇒ 8__+ 10H₂O
Required
product compound
Solution
In the combustion of hydrocarbons there can be 2 kinds of products
If there is excess Oxygen, you will get Carbon dioxide(CO₂) and water in the product
If Oxygen is low, you'll get Carbon monoxide(CO) and water
Or in other ways, we can use the principle of the law of conservation of mass which is also related to the number of atoms in the reactants and in the products
if we look at the reaction above, there are C atoms on the left (reactants), so that in the product there will also be C atoms with the same number of C atoms on the left
2C₄H₁₀ + 13O₂⇒ 8CO₂+ 10H₂O
Answer:
The specific rotation of D is 11.60° mL/g dm
Explanation:
Given that:
The path length (l) = 1 dm
Observed rotation (∝) = + 0.27°
Molarity = 0.175 M
Molar mass = 133.0 g/mol
Concentration in (g/mL) = 0.175 mol/L × 133.0 g/mol
Concentration in (g/mL) = 23.275 g/L
Since 1 L = 1000 mL
Concentration in (g/mL) = 0.023275 g/mL
The specific rotation [∝] = ∝/(1×c)
= 0.27°/( 1 dm × 0.023275 g/mL
)
= 11.60° mL/g dm
Thus, the specific rotation of D is 11.60° mL/g dm
What I think is the charge of nucleus is the proton+neutron
<u>Answer:</u> The equilibrium concentration of CO is 0.243 atm
<u>Explanation:</u>
We are given:
Initial partial pressure of carbon dioxide = 0.902 atm
As, carbon dioxide is present initially. This means that the reaction is proceeding backwards.
For the given chemical equation:
<u>Initial:</u> 0.902
<u>At eqllm:</u> 3x (0.902-3x)
The expression of 
for above equation follows:
We are given:
Putting values in above equation, we get:
So, equilibrium concentration of CO = 3x = (3 × 0.0810) = 0.243atm[/tex]
Hence, the equilibrium concentration of CO is 0.243 atm
