A. High intermolecular forces of attraction. If there are high intermolecular forces, the molecules will need large energies to escape into the liquid. The substance will nave a high melting point.
The other options are <em>incorrect </em>because they are <em>weak force</em>s. They would cause <em>low melting points</em>.
Carbon dioxide and oxygen are removed from the air.
Explanation:
When air is passed through aqueous sodium hydroxide solution the carbon dioxide is removed from the air.
First the carbon dioxide will dissolve and react with water to form carbonic acid ( H₂CO₃) :
CO₂ + H₂O → H₂CO₃
The the carbonic acid will react with sodium hydroxide to form sodium carbonate (Na₂CO₃):
H₂CO₃ + 2 NaOH → Na₂CO₃ + 2 H₂O
After this by passing the air over heated cooper the oxygen is removed.
2 Cu + O₂ → 2 CuO
Learn more about:
neutralization reaction
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Answer : The oxidation state of Mg in Mg(s) is (0).
Explanation :
Oxidation number or oxidation state : It represent the number of electrons lost or gained by the atoms of an element in a compound.
Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.
Rules for Oxidation Numbers are :
The oxidation number of a free element is always zero.
The oxidation number of a monatomic ion equals the charge of the ion.
The oxidation number of Hydrogen (H) is +1, but it is -1 in when combined with less electronegative elements.
The oxidation number of oxygen (O) in compounds is usually -2.
The oxidation number of a Group 17 element in a binary compound is -1.
The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.
The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.
The given chemical reaction is:
In the given reaction, the oxidation state of Mg in Mg(s) is (0) because it is a free element and the oxidation state of Mg in 
is (+2).
Hence, the oxidation state of Mg in Mg(s) is (0).
Answer:
y1 = 0.3162
y2 = 0.6838
Explanation:
ok let us begin,
first we would be defining the parameters;
at 25°C;
1-propanol P1° = 20.90 Torr
2-propanol P2° = 45.2 Torr
From Raoults law:
P(1-propanol) = P⁰ × X(1-propanol)
P(1-propanol) = 20.9 torr × 0.45 = 9.405
P(1-propanol) = 9.405 torr
Also P(2-propanol) = P⁰ × X(2-propanol)
P(2-propanol) = 45.2 torr × 0.45
P(2-propanol) = 20.34 torr
but the total pressure = sum of individual pressures
total pressure = 9.405 + 20.34
total pressure = 29.745 torr
given that y1 and y2 represent the mole fraction of each in the vapor phase
y1 = P1 / total pressure
y1 = 9.405/29.745
y1 = 0.3162
Since y1 + y2 = 1
y2 = 1 - y1
∴ y2 = 1 - 0.3162
y2 = 0.6838
cheers, i hope this helps.
As the gas cools it condenses and becomes a liquid its atoms also become smaller
