Answer
is: 0.375 moles are present in 8.4 liters of nitrous oxide at stp.
V(N₂O) = 8.4 L.
V(N₂O) =
n(N₂O) · Vm.
Vm = 22,4 L/mol.<span>
n</span>(N₂O) = V(N₂O) ÷ Vm.
n(N₂O) = 8.4 L ÷ 22.4 L/mol.
n(N₂O) = 0.375 mol.<span>
Vm - molare volume on STP.</span>
An independent variable is the variable you are changing in order to measure the dependent variable, which is what you are measuring.
In this example, the
independent variable: chemicals in solution
dependent variable: temperature of solution
Is true. Nitrogen gas behaves more like an ideal gas as the
temperature increases. Under normal conditions such as normal pressure and temperature
conditions , most real gases behave qualitatively as an ideal gas. Many
gases such as air , nitrogen , oxygen ,hydrogen , noble gases , and some heavy
gases such as carbon dioxide can be treated as ideal gases within a reasonable tolerance. Generally,
the removal of ideal gas conditions tends to be lower at higher temperatures and lower density (that is at lower pressure ), since the work made by the intermolecular
forces is less important compared to the kinetic energy<span> of the particles, and the size of the molecules is less important
compared to the empty space between them. </span><span>The ideal gas model
tends to fail at lower temperatures or at high pressures, when intermolecular
forces and intermolecular size are important.</span>
Answer:
C) As a gas is heated, the pressure decreases.
Explanation:
From the choices given, the statement that "as a gas is heated, the pressure decreases is false".
When gases are heated, their molecules expands and the volume of the gas increases. In a fixed container, the pressure of the gases will also increases.
- Gases lack internal cohesion and very weak to no intermolecular forces binding them together.
- When they are subjected to heat, they gain more energy(kinetic energy) which causes them to begin to spread out.
- Thus, they take up even more space allowing volume to increase appreciably.
Answer:
9.8g
Explanation:
Using periodic table find molar mass of C and H:
C=12.01g/mol
H=1.008g/mol
Molar mass of C2H4=(12.01)2+(1.008)4=28.03g/mol
Using molar mass times moles of the chemical to find the mass in 0.35 moles of c2h4:
28.03 x 0.35=9.8grams