Respuesta :
Answer:
The element X is chlorine.
The element M is Ytterium.
Ytterium (III) chloride is the correct name for [tex]YCl_3[/tex]
Explanation:
Molecules of [tex]X_2, N = 8.92\times 10^{20} [/tex]
Moles of [tex]X_2, n= ?[/tex]
[tex]N=n\times N_A[/tex]
[tex]n=\frac{N}{N_A}=\frac{8.92\times 10^{20}}{6.022\times 10^{23} mol^{-1}}[/tex]
n = 0.001481 mol
Mass of [tex]X_2, m=0.105 g[/tex]
Molar mass of [tex]X_2=M[/tex]
[tex]m=M\times n[/tex]
[tex]M=\frac{m}{n}=\frac{0.105 g}{0.001481 mol}[/tex]
M = 70.87 g/mol
Atomic mass of X = [tex]\frac{70.87 g/mol}{2}=35.435 g/mol[/tex]
X is a chlorine atom.
The compound [tex]MX_3[/tex] consists of 54.47% X by mass.
M' = Molar mass of compound [tex]MX_3[/tex]
Percentage of X in compound = 54.47%
[tex]54.47\%=\frac{3\times 35.435 g/mol}{M'}\times 100[/tex]
M' = 195.16 g/mol
Atomic mass of M = a
[tex]195.16 g/mol=a+3\times 35.435 g/mol[/tex]
[tex]a = 195.16 g/mol-3\times 35.435 g/mol[/tex]
a = 88.86 g./mol
The element M is Ytterium.
Ytterium (III) chloride is the correct name for [tex]YCl_3[/tex]
The element X is Chlorine and the element M is Ytterium.
Ytterium Chloride is the correct name for [tex]MX_3[/tex] that is [tex]YCl_3[/tex]
Number of molecules of [tex]X_2={8.92*10^{20}[/tex]
Number of moles of X:
[tex]n=\frac{8.92*10^{20}}{6.023*10^{23}}[/tex]
n = 0.001481 mol = 0.105g
Molar mass of [tex]X_2[/tex] molecule
[tex]M= 0.105/0.001481\\\\M=70.87g/mol[/tex]
Therefore molar mass of X atom = molar mass of [tex]X_2[/tex]/2
m = 35.435g/mol
X is a chlorine atom.
Given that the compound consists of 54.47% X by mass.
Let M' = Molar mass of compound [tex]MX_3[/tex]
Percentage of X in compound = 54.47%
54.47 = {amount of X}/{mola mass of compound}*100
[tex]54.47 = \frac{3*35.435}{M_{'} }*100\\\\M^{'}= 195.16g/mol[/tex]
let Atomic mass of M = A
A = 195.16 - 3 × 35.435
A = 88.86 g/mol
The element M is Ytterium.
Therefore the compound [tex]MX_3[/tex] is Ytterium Chloride [tex]YCl_3[/tex]
Learn more:
https://brainly.com/question/1855861
