Answer:
Density = mass/volume
= 44/22.4
= 1.96 gram/liter
The density of the Carbon Dioxide at S.T.P. (Standard Temperature and Volume) is 1.96 gram/liter.
Physical change - No change of matter in this phase
chemical change - All types of phase change occur here
First, consider the steps to heat the sample from 209 K to 367K.
1) Heating in liquid state from 209 K to 239.82 K
2) Vaporaizing at 239.82 K
3) Heating in gaseous state from 239.82 K to 367 K.
Second, calculate the amount of heat required for each step.
1) Liquid heating
Ammonia = NH3 => molar mass = 14.0 g/mol + 3*1g/mol = 17g/mol
=> number of moles = 12.62 g / 17 g/mol = 0.742 mol
Heat1 = #moles * heat capacity * ΔT
Heat1 = 0.742 mol * 80.8 J/mol*K * (239.82K - 209K) = 1,847.77 J
2) Vaporization
Heat2 = # moles * H vap
Heat2 = 0.742 mol * 23.33 kJ/mol = 17.31 kJ = 17310 J
3) Vapor heating
Heat3 = #moles * heat capacity * ΔT
Heat3 = 0.742 mol * 35.06 J / (mol*K) * (367K - 239.82K) = 3,308.53 J
Third, add up the heats for every steps:
Total heat = 1,847.77 J + 17,310 J + 3,308.53 J = 22,466.3 J
Fourth, divide the total heat by the heat rate:
Time = 22,466.3 J / (6000.0 J/min) = 3.7 min
Answer: 3.7 min
Answer:
snow, or freezing rain?
Explanation:
because freezing rain can occur then, but snow is more common.
Answer:
0.7457 g is the mass of the helium gas.
Explanation:
Given:
Pressure = 3.04 atm
Temperature = 25.0 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (25.0 + 273.15) K = 298.15 K
Volume = 1.50 L
Using ideal gas equation as:

where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
3.04 atm × 1.50 L = n × 0.0821 L.atm/K.mol × 298.15 K
<u>⇒n = 0.1863 moles</u>
Molar mass of helium = 4.0026 g/mol
The formula for the calculation of moles is shown below:
Thus,

<u>0.7457 g is the mass of the helium gas. </u>