Rutherford was one of the early scientists who worked on the atomic model. Before his discovery of the nucleus, the widely accepted theory was J.J Thomson's Plum Pudding Model. In this model, all the protons, electrons and neutrons are in the nucleus. But the electrons are more in number such that the electrons act as the 'pudding' and the proton and nucleus the 'plum'. This was Rutherford's hypothesis in his gold foil experiment. In order to test the Plum Pudding model, he hypothesized that when a beam of light is aimed at the atom, it would not diffract because the charges in the nucleus are well-distributed. However, his experiment disproved Thomson's model. Some light indeed passed through but a few was diffracted back to the source. He concluded that this was because there is a dense mass inside the atom called nucleus. Thus, from there on, he proposed the model that the electrons are orbiting around the nucleus.
Answer:
B
Explanation:
Having the same electronegativity means that the both elements will form a nonpolar bond. The shared electrons of the bind will be found equidistant from the nucleus of the two atoms because none has an unusual ability to attract the electrons of a bond towards itself.
The formula that would represent an ionic compound that is composed of calcium and iodide ions is CaI2
Explanation
Ionic compound CaI₂ is formed when Calcium form cation ( <em>a positively charged ion</em>) by losing 2 electrons while two iodine atoms form anion ( <em>a negatively charged ion</em>) by gaining one electron each.
When writing down formula of ionic compound, the formula of cation is written first followed by anion formula. therefore Ca is written first followed by I.
The numeric subscript 2 after I(iodine) indicate that 2 atoms of iodine are involved in bonding.
Answer:
1.62
Explanation:
From the given information:
number of moles of benzamide 
= 0.58 mole
The molality = 
= 0.6837
Using the formula:
where;
dT = freezing point = 27
l = Van't Hoff factor = 1
kf = freezing constant of the solvent
∴
2.7 °C = 1 × kf × 0.6837 m
kf = 2.7 °C/ 0.6837m
kf = 3.949 °C/m
number of moles of NH4Cl = 
= 1.316 mol
The molality = 
= 1.5484
Thus;
the above kf value is used in determining the Van't Hoff factor for NH4Cl
i.e.
9.9 = l × 3.949 × 1.5484 m
l = 1.62
Answer:
[HI] = 0.7126 M
Explanation:
Step 1: Data given
Kc = 54.3
Temperature = 703 K
Initial concentration of H2 and I2 = 0.453 M
Step 2: the balanced equation
H2 + I2 ⇆ 2HI
Step 3: The initial concentration
[H2] = 0.453 M
[I2] = 0.453 M
[HI] = 0 M
Step 4: The concentration at equilibrium
[H2] = 0.453 - X
[I2] = 0.453 - X
[HI] = 2X
Step 5: Calculate Kc
Kc = [Hi]² / [H2][I2]
54.3 = 4x² / (0.453 - X(0.453-X)
X = 0.3563
[H2] = 0.453 - 0.3563 = 0.0967 M
[I2] = 0.453 - 0.3563 = 0.0967 M
[HI] = 2X = 2*0.3563 = 0.7126 M
