Answer:
1.1 × 10⁻⁴ M
Explanation:
Let's consider the following double displacement reaction.
CuCl₂(aq) + 2 AgNO₃(aq) → 2 AgCl(s)+ Cu(NO₃)₂(aq)
We can establish the following relations:
- The molar mass of AgCl is 143.32 g/mol.
- The molar ratio of AgCl to CuCl₂ is 2:1
The moles of CuCl₂ that reacted to produce 7.7 mg of AgCl are:

The molarity of CuCl₂ is:

Answer:
1 mole of platinum
Explanation:
To obtain the number of mole(s) of platinum present, we need to determine the empirical formula for the compound.
The empirical formula for the compound can be obtained as follow:
Platinum (Pt) = 117.4 g
Carbon (C) = 28.91 g
Nitrogen (N) = 33.71 g
Divide by their molar mass
Pt = 117.4 / 195 = 0.602
C = 28.91 / 12 = 2.409
N = 33.71 / 14 = 2.408
Divide by the smallest
Pt = 0.602 / 0.602 = 1
C = 2.409 / 0.602 = 4
N = 2.408 / 0.602 = 4
The empirical formula for the compound is PtC₄N₄ => Pt(CN)₄
From the formula of the compound (i.e Pt(CN)₄), we can see clearly that the compound contains 1 mole of platinum.
a substance that increases the rate of a chemical reaction without itself undergoing any permanent chemical change.
- a person or thing that precipitates an event.