The reaction between halogens and alkanes is very rapid. true false

A chemist prepares a solution of copper(II) fluoride by measuring out of copper(II) fluoride into a volumetric flask and filling

Simora [160]

The question is incomplete, here is the complete question.

A chemist prepares a solution of copper(II) fluoride by measuring out 0.0498 g of copper(II) fluoride into a 100.0mL volumetric flask and filling the flask to the mark with water.

Calculate the concentration in mol/L of the chemist's copper(II) fluoride solution. Round your answer to 3 significant digits.

Answer: The concentration of copper fluoride in the solution is $4.90\times 10^{-3}mol/L$

Explanation:

To calculate the molarity of solute, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Given mass of copper (II) fluoride = 0.0498 g

Molar mass of copper (II) fluoride = 101.54 g/mol

Volume of solution = 100.0 mL

Putting values in above equation, we get:

$\text{Molarity of copper (II) fluoride)=\frac{0.0498\times 1000}{101.54\times 100.0}\\\\\text{Molarity of copper (II) fluoride}=4.90\times 10^{-3}mol/L$

Hence, the concentration of copper fluoride in the solution is $4.90\times 10^{-3}mol/L$

4 0

3 years ago

a ______ consumer is a heterotroph that directly eats an autotroph. A primary B. Quaterany C tertiary D secondary

Softa [21]

Primary consumer just remember the “primary” (first) consumer so the one that eats a heterotroph

6 0

3 years ago

Which is the Net Ionic Equation for:

arlik [135]

Answer: $2Fe^{3+}(aq)+3Mg(s)\rightarrow 3Mg^{2+}(aq)+2Fe(s)$

Explanation:

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and are present in an aqueous state and represented in the form of ions.

Net ionic equation : In the net ionic equations, we do not not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The complete balanced ionic equation will be:

$2Fe^{3+}(aq)+6Cl^-(aq)+3Mg(s)\rightarrow 3Mg^{2+}(aq)+6Cl^-(aq)+2Fe(s)$