Exactly 3.705kg of substance Y are needed to neutralise 100 moles of HCL aq. What could be substance Y ? A. Mgcl2 and Caco3 B. M

Answer:

(i). The elements that tend to create a cation

(C). calcium, (D). barium, (E). cesium, (F). rubidium, (G). lead and (H). scandium

(ii). The elements that always form ionic compounds in a 1:2 ratio with halogens

(B). lead, (G) barium and (I). calcium

Explanation:

Considering that,

The following elements list,

A. oxygen B. chlorine C. calcium D. barium E. cesium F. rubidium G. lead H. scandium I. phosphorous

It is understood that,

A cation is defined as an element that can easily donate an electron and transforms into a positive ion.

All metal elements have the capability to form cations.

We are tasked with identifying the elements likely to produce a cation

Based on the provided list

(C). calcium

(D). barium

(E). cesium

(F). rubidium

(G). lead

(H). scandium

(II). In the next list,

A. phosphorous B. lead C. chlorine D. cesium E. rubidium F. scandium G. barium H. oxygen I. calcium

It is known that,

All metal ions listed can create ionic compounds with halogens at a 1:2 ratio because all given metals can form +2 charges.

This indicates the necessity to find those elements that will invariably form ionic compounds in a 1:2 ratio with halogens

Using the provided list.

(B). lead

(G) barium

(I). calcium

Thus, (i). The elements that can create a cation

(C). calcium, (D). barium, (E). cesium, (F). rubidium, (G). lead and (H). scandium

(ii). The elements that will consistently form ionic compounds in a 1:2 ratio with the halogens

(B). lead, (G) barium and (I). calcium

The chemical reaction can be expressed with this balanced equation:
 <span>SrBr2(aq) + 2AgNO3(aq) → Sr(NO3)2(aq) + 2AgBr(s) 
</span>
Using the periodic table:
Silver has a mass of 108 grams
Bromine weighs 80 grams

Thus, the molar mass of silver bromide amounts to 188 grams (108 + 80).

To find the number of moles, we use the formula: number of moles = mass / molar mass
The moles of precipitate formed can be calculated as 3.491/188 = 0.018 moles.

According to the balanced equation:
1 mole of strontium bromide gives rise to 2 moles of silver bromide. Therefore, we can find the moles of strontium bromide required to create 0.018 moles of silver bromide by cross-multiplying as follows:
amount of<span>strontium bromide = (0.018x1) / 2 = 9.28 x 10^-3 moles</span>

Answer:

The mass percentage of GaBr₃ in the solid mixture is 30.2 %.

Explanation:

GaBr₃(aq) + 3 AgNO₃(aq) ⟶ 3 AgBr(s) + Ga(NO₃)₃(aq)

To find the molar mass, we have: MW GaBr₃ = 309.4 g/mol and MW AgBr = 187.8 g/mol.

Using these values, we note that 187.8 g of AgBr corresponds to 1 mol.

If 0.368 g of AgBr yields x moles, we calculate x to be 2.0 x 10⁻³ mol AgBr.

Based on the stoichiometry, 1 mol of GaBr₃ produces 3 mol of AgBr, giving us:

y ___________ 2.0 x 10⁻³ mol AgBr

Thus, y equals 6.7 x 10⁻⁴ mol GaBr₃.

Additionally, 1 mol of GaBr₃ has a mass of 309.4 g.

Consequently, 6.7 x 10⁻⁴ mol of GaBr₃ corresponds to w grams of GaBr₃, yielding w = 0.206 g GaBr₃.

Relating this to the total mass:

0.6813 g _____ 100%

0.206 g _____ z

Therefore, z is calculated as 30.2 %.

Answer:

Explanation:

Chlorine has an electron configuration of 2, 8, 7

Within the n = 3 shell, it has 7 electrons consisting of 2 in the s subshell and 5 in the p subshell. However, one of the p electrons shifts to the d orbital, altering the electronic configuration  to be as follows

 = 7

These orbitals such as sp³d hybridize, creating 7 degenerate orbitals where two orbitals are filled with electron pairs and three remain singly occupied by electrons (unpaired electrons).