Cause of iron changing to rust when it reacts with water
1 answer:
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Answer:
The density of the ideal gas is directly proportional to its molar mass.
Explanation:
Density is a scalar quantity that is denoted by the symbol ρ (rho). It is defined as the ratio of the mass (m) of the given sample and the total volume (V) of the sample.
......equation (1)
According to the ideal gas law for ideal gas:
......equation (2)
Here, V is the volume of gas, P is the pressure of gas, T is the absolute temperature, R is Gas constant and n is the number of moles of gas
As we know,
The number of moles:
where m is the given mass of gas and M is the molar mass of the gas
So equation (2) can be written as:
⇒
⇒ ......equation (3)
Now from equation (1) and (3), we get
⇒ Density of an ideal gas:
⇒ <em>Density of an ideal gas: ρ ∝ molar mass of gas: M</em>
<u>Therefore, the density of the ideal gas is directly proportional to its molar mass. </u>
If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl alcohol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785g/cm^3 .
will all the alcohol evaporate? or none at all?
Answer:
Yes, all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.
Explanation:
Given that:
The volume of alcohol which is placed in a small laboratory = 1.0 L
Vapor pressure of ethyl alcohol at 25 ° C = 59 mmHg
Converting 59 mmHg to atm ; since 1 atm = 760 mmHg;
Then, we have:
=
= 0.078 atm
Temperature = 25 ° C
= ( 25 + 273 K)
= 298 K.
Density of the ethanol = 0.785 g/cm³
The volume of laboratory = l × b × h
= 3.0 m × 2.0 m × 2.5 m
= 15 m³
Converting the volume of laboratory to liter;
since 1 m³ = 100 L; Then, we have:
15 × 1000 = 15,000 L
Using ideal gas equation to determine the moles of ethanol in vapor phase; we have:
PV = nRT
Making n the subject of the formula; we have:
n = 47. 88 mol of ethanol
Moles of ethanol in 1.0 L bottle can be calculated as follows:
Since numbers of moles =
and mass = density × vollume
Then; we can say ;
number of moles =
number of moles =
number of moles =
number of moles = 17.039 mol
Thus , all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.