Based on the Lewis/electron dot representation of the two atoms, predict the ratio of metal
cationic (+) atom to nonmetal anionic (-) atom in the compound

Lewis dot structure is a diagrammatic representation of valence electrons in an atom of a molecule.

The ratio of the cations to anions in the given compound is 1:3.

In the given figure, Aluminum has 3 electrons, which can be donated to iodine. Aluminum has atomic number 13, this means that if the element donates three electrons then it will attain a stable electronic configuration.

In the Iodine atom, each iodine needs only one electron to complete its octet. Thus, three atoms of iodine will be required to gain electrons from aluminum.

Thus, the ratio cation to the anion is 1:3.

To know more about lewis dot structure, refer to the following link:

link

A. 1:3

Explanation:

If we look at the ions shown in the image attached to the question, we will notice that we have aluminum (Al^3+), a trivalent ion combining with the iodide ion (I^-).

Aluminum can easily give out its three outermost electrons to three atoms of iodine. If aluminum gives out its three electrons, it achieves the stable octet structure. Iodine atoms have seven electrons in their outermost shell. They only need one more electrons to complete their octet. This one electron can be gotten by the combination of three iodine atoms with one atom of aluminum. One electron each is transferred from the aluminum atom to each iodine atom to form AlI3 with a ratio of 1:3.

where's the diagram?

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol