Answer:
This would support Dalton's postulates that proposed the atoms are indivisible because no small particles are involved.
Explanation:
Experiment using the gas discharge tube by J.J Thomson led to the discovery of cathode rays which are now known as electrons.
Primarily, Thomson's experiment led to the discovery of cathode rays, electrons, as subatomic particles.
If the size of the atoms observed at the cathode is the same as that of the rays,we can conclude that the particles of the rays are the simplest form of matter we can have. This would suggest that the atom is indeed the smallest indivisible particle of a matter according to Dalton.
Answer :
The correct answer is %IC = 10 % and bond is covalent bond with slight polarity.
<u>Percent Ionic Character :</u>
It is defined as percent of ionic character present in a polar covalent bond . The formula of % ionic character (%IC) is given as follows :
Where Xa = Electronegativity of A atom and Xb = Electronegativity of B atom
Given : Molecule is TiAl₃
Electronegativity of Ti = 2.0
Electronegativity of Al = 1.6 ( From image shared )
Plug the value in above formula :
Value of e⁻¹ = 0.90
Percent ionic character = 1 - 0.90 * 100
Percent Ionic character = 10 %
<u>Since the % IC is 10 % , which is very less comparatively , hence the bond is covalent and very less polar .</u>
Answer:
the change is evaporation
Explanation:
the water heats up at the surface of the water and evaporates
Answer:
Option (A). The mass of the rusted nail equals the mass of iron and the oxygen from the air it reacted with to form the rust.
When the amount of heat gained = the amount of heat loss
so, M*C*ΔTloses = M*C* ΔT gained
when here the water is gained heat as the Ti = 25°C and Tf= 28°C so it gains more heat.
∴( M * C * ΔT )W = (M*C*ΔT) Al
when Mw is the mass of water = 100 g
and C the specific heat capacity of water = 4.18
and ΔT the change in temperature for water= 28-25 = 3 ° C
and ΔT the change in temperature for Al = 100-28= 72°C
and M Al is the mass of Al block
C is the specific heat capacity of the block = 0.9
so by substitution:
100 g * 4.18*3 = M Al * 0.9*72
∴ the mass of Al block is = 100 g *4.18 / 0.9*72
= 19.35 g
