Answer:
A
Explanation:
In all the other options either Carbon has more than 4 atoms attached or less than 4.
Carbon can not form more than four bonds. It can only share 4 electrons .
In other option C, No H-atom is linked in the right most Carbon.
whereas according to the definition of Hydrocarbon it should have been a Hydrogen atom.
Answer:
Substances 1 and 2
Explanation:
an element only has 1 kind of atoms :3
The new volume be if the pressure inside the balloon was reduced to 58kpa is 2.58 L.
<h3>What is ideal gas equation?</h3>
Ideal gas equation PV=nRT gives idea about the behavior of gases at different conditions and for this question the equation becomes:
P₁V₁ = P₂V₂, where
P₁ = initial pressure = 115.3 kpa
V₁ = initial volume = 1.3 L
P₂ = final pressure = 58 kpa
V₂ = final volume = ?
On putting all these values in the above equation, we get
V₂ = (115.3)(1.3) / (58) = 2.58 L
Hence resultant volume of gas is 2.58 L.
To know more about ideal gas equation, visit the below link:
brainly.com/question/18909295
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Answer:
you can't change it mate I am so sorry
BTW can you help me in my latest questions it's not related to studies it's a problem I am having with my laptop that need to get cured as I have exam tomorrow
Answer:

Explanation:
Balanced equation: CO(g) + H₂O(g) ⟶ CO₂(g) + H₂(g)
We can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products

(a) Enthalpies of formation of reactants and products

(b) Total enthalpies of reactants and products

(c) Enthalpy of reaction