Answer:
a. 4,00L
b. 16,00L
c. 12,31L
Explanation:
Avogadro's law says:
a. If initial conditions are 2,30mol and 8,00L and you lose one-half of atoms, that means you have 1,15mol:
<em>V₂ = 4,00L</em>
b. If initial conditions are 2,30mol and 8,00L and you add 2,30mol, that means you have 4,60mol:
<em>V₂ = 16,00L</em>
c. 25,0g of Ne are:
25,0g × (1mol / 20,1797g) = 1,24 moles of Ne. That means you have 2,30mol - 1,24mol = 3,54mol of Ne
<em>V₂ = 12,31L</em>
I hope it helps!
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Answer:
Atoms making liquids have less attraction than solids, but more than gases
Explanation:
The attraction between atoms in different molecules in a solid is very strong due to strong intermolecular forces present in a solid. However, such intermolecular forces are weaker in liquids than in solids.
This implies that the solid has higher intermolecular forces of attraction compared to gases and liquids. Based on the negligible degree of intermolecular forces between them, a gas has the weakest intermolecular forces hence the atom has very minimal interaction between them.
20 g O2 x 1 mol O2/32 g O = 0.625 mol O2
You should always study. you can be above your class. thats what i do.. search google for ur course. they can tell u some key points for it. u should even study what you know and want to know as well.
