Answer:
The classification and illustrations are attached in the drawing.
Explanation:
It is possible to identify the pure substance observing the figure, since it is the only one that has 2 joined atoms (purple and blue) which forms a single compound.
On the other hand, the homogeneous mixture is identified by noting that its atoms are more united with respect to the heterogeneous mixture, highlighting that in homogenous mixtures the atoms, elements or substances are not visible to the naked eye and are in a single phase, instead in the heterogeneous mixture if they can be differentiated.
Answer:
2.8 * 10^(-6) / 1.4 * 10^(-2)=
2* 10^(-8)
We will use Arrehenius equation
lnK = lnA -( Ea / RT)
R = gas constant = 8.314 J / mol K
T = temperature = 25 C = 298 K
A = frequency factor
ln A = ln (1.5×10 ^11) = 25.73
Ea = activation energy = 56.9 kj/mol = 56900 J / mol
lnK = 25.73 - (56900 / 8.314 X 298) = 2.76
Taking antilog
K = 15.8
Air is a mixture, Its constituens can be seperated
Proton:
Positive
Found in Nucleus
Mass of 1 AMU
Neutron:
Neutral
Found in Nucleus
Mass of 1 AMU
Electron:
Negative
Found in orbitals
Mass of 0 AMU
Please mark Brainliest if this helped you! :)