a) Gas particles have most of their mass concentrated in the nucleus of the atom.
b) The moving particles undergo perfectly elastic collisions with the walls of the container.
c) The forces of attraction and repulsion between the particles are insignificant.
d) The average kinetic energy of the particles is directly proportional to the absolute temperature.
e) All of the above are postulates of the kinetic molecular theory.
Gas particles have most of their mass concentrated in the nucleus of the atom.
Answer: Option A.
<u>Explanation:</u>
Kinetic Molecular Theory expresses that gas particles are in consistent movement and show flawlessly versatile crashes. Motor Molecular Theory can be utilized to clarify both Charles' and Boyle's Laws. The normal active vitality of an assortment of gas particles is straightforwardly corresponding to total temperature as it were.
The kinetic theory of gases is a significantly critical, however straightforward model of the thermodynamic conduct of gases with which numerous important ideas of thermodynamics were built up.
Answer: A
Explanation: cytokinesis is in the M phase.... Mitosis is followed by cytokinesis (cell separation)
The best description of Ernest Rutherford's experiment is letter C. The positively charged particles were fired through a gold foil.
Answer:
136.63 °C
Explanation:
ΔTb=Tb solution - Tb pure
Where; Tb pure = 133.60°C
molar mass of solute = 121.14 g/mol
number of moles of solute; 52.2g/121.14 g/mol = 0.431 moles
molality = 0.431 moles/350 * 10^-3 = 1.23 molal
Then;
ΔTb = Kb * m * i
Kb = 2.46°C kg mol^-1
m = 1.23 molal
i = 1
ΔTb = 2.46 * 1.23 * 1
ΔTb = 3.03 °C
Hence;
Tb solution = ΔTb + Tb pure
Tb solution = 3.03 °C + 133.60°C
Tb solution = 136.63 °C
Answer:
Because we need to dispense 4.7 mL, the volume reading in the pipet is the 5.3 mL line.
Explanation:
First we use C₁V₁=C₂V₂ in order to <u>calculate the required volume of concentrated HCl</u> (V₁):
12.85 M * V₁ = 0.600 M * 100 mL
V₁ = 4.7 mL
<u>So we need to dispense 4.7 mL of the concentrated HCl solution</u>. The mark in the pipet that would contain that volume would be 10.0 - 4.7 = 5.3 mL
