M+ is an unknown metal cation with a plus one charge. As it dissolves the chloride of the unknown metal, MCL, in enough water to make 100 mL of solution. The student the next is a solution with excess AgNO3 solution, causing a GCL to precipitate. The student collects a precipitate by filtration

The question is incomplete; the complete question is;

M+ is an unknown metal cation with a +1 charge. A student dissolves the chloride of the

unknown metal, MCl, in enough water to make 100.0 mL of solution. The student then mixes the

solution with excess AgNO3 solution, causing AgCl to precipitate.

The student collects the precipitate by filtration, dries it, and records the data shown

below. (The molar mass of AgCl is 143 g/mol.)

Mass of unknown chloride, MCl 0.74 g

Mass of filter paper 0.80 g

Mass of filter paper plus AgCl precipitate 2.23 g

1. What is the identity of the metal chloride?

(A) NaCl

(B) KCl

(C) CuCl

(D) LiCl

KCl

Explanation:

Mass of paper= 0.80 g

Mass of paper and AgCl=2.23

Hence mass AgCl = 2.23 - 0.80 = 1.43g AgCl

Number of moles of AgCl= 1.43g AgCl / 143g/mol AgCl = 0.01 moles AgCl

Since Ag+ and Cl- are in a mole ratio of 1:1 so:

  number   moles Ag+ = number moles Cl-,

Thus Ag+ has 0.01 moles and Cl- has 0.01 moles

- MCl is also in a 1:1 ratio

Hence;

 number of moles of moles Cl- =number of moles of moles M+  = 0.01 moles of M+ and Cl-

Thus;

0.01 Cl- = x/35.45 = 0.3545g of Cl-

mass of MCl = 0.74g

Mass of M+ =0.74g MCl - 0.3545g Cl- = 0.3955g  of M+

Number of moles of M+ =0.3955g M+/x =0.010 mol M+

x= 39.55g of M+

K+ has a molar mass of approximately 39.10

Hence KCl is the answer

the temperature of the gas, in k, not celsius.

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer