He made a new approach when he discovered magnetism and electricity.
The mole fraction of a gaseous compound is equal to the ratio of the vapor pressure of the compound to the total pressure of the vessel. In this case, teh vapor pressure of tehe non-electrolyte is 760 - 745 = 15 mmHg. hence the mole fraction is 15 mm Hg / 760 mmHg equal to 0.0197
Answer:
Here's what I get
Explanation:
1. Fluorine
a. Each fluorine atom has seven valence electrons.
b. Fluorine is a nonmetal.
c. The combination of these atoms creates a covalent bond.
d. The atoms share a pair of valence electrons.
e. In the ionic bond, an electron moved from one atom to the other one.
f. The molecule forms a straight line.
2. Oxygen
a. Each oxygen atom has six valence electrons.
b. Oxygen is a nonmetal.
c. The combination of these atoms creates a covalent bond.
d. The atoms share two pairs of valence electrons.
e. The bond has four electrons instead of two.
f. The molecule forms a straight line.
<u>Answer:</u> The weight fraction of ethanol in the mixture is
<u>Explanation:</u>
We are given:
Mass of water = 1 kg = 2.205 lb (Conversion factor: 1 kg = 2.205 lb)
Mass of ethanol = 1.9 lb
Mass of ethyl acetate = 4.6 lb
Mass of mixture = [2.205 + 1.9 + 4.6] = 8.705 lb
To calculate the percentage composition of ethanol in mixture, we use the equation:
Mass of mixture = 8.705 lb
Mass of ethanol = 1.9 lb
Putting values in above equation, we get:
Hence, the weight fraction of ethanol in the mixture is 21.8 %
In order to find the two statements, we must first define what the enthalpy of formation and the enthalpy of reaction mean.
Enthalpy of formation:
The change in enthalpy when one mole of substance is formed from its constituent elemetns at standard state.
Enthalpy of reaction:
The change in enthalpy when a reaction occurs and the reactants and products are in their standard states.
Now, we check the statements. The true ones are:
The Hrxn for C(s) + O₂(g) → CO₂(g) is the same as Hf for CO₂
This is true because the formation of carbon dioxide requires carbon and oxygen in their standard states.
The Hf for Br₂<span>(l) is 0 kJ/mol by definition.
Because the bromine is present in its standard state, the enthalpy of formation is 0.
</span><span>The Hrxn for the reaction 1.5H</span>₂<span>(g) + 0.5N</span>₂<span>(g) </span>→ <span>NH</span>₃<span>(g) is the same as the Hf for NH</span>₃<span>(g)
The reactants and products are present in their standard state, and the reaction is the same as the one occurring during the formation of ammonia.
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