Answer:
0.3936 J/gC
Explanation:
using the formula: q=mcΔt
q= 7032J
m=812g
ΔT = 22C
plug in and solve:
7032=(812)(c)(22)
c=7032/(812)(22)
c=0.39 J/gC
An oxidation-reduction (redox<span>) </span>reaction<span> is a type of chemical </span>reaction<span> that involves a transfer of electrons between two species. An oxidation-reduction </span>reaction<span> is any chemical </span>reaction<span> in which the oxidation number of a molecule, atom, or ion changes by gaining or losing an electron.</span>
Answer:
The final step is to <u>REPORT</u>!!!
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.
<u>Answer:</u> The concentration of copper fluoride in the solution is 
<u>Explanation:</u>
To calculate the molarity of solute, we use the equation:

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:

Hence, the concentration of copper fluoride in the solution is 
chemical reactions can release heat (a)