Answer:
B. A precipitate will form since Q > Ksp for calcium oxalate
Explanation:
Ksp of CaC₂O₄ is:
CaC₂O₄(s) ⇄ Ca²⁺ + C₂O₄²⁻
Where Ksp is defined as the product of concentrations of Ca²⁺ and C₂O₄²⁻ in equilibrium:
Ksp = [Ca²⁺][C₂O₄²⁻] = 2.27x10⁻⁹
In the solution, the concentration of calcium ion is 3.5x10⁻⁴M and concentration of oxalate ion is 2.33x10⁻⁴M.
Replacing in Ksp formula:
[3.5x10⁻⁴M][2.33x10⁻⁴M] = 8.155x10⁻⁸. This value is reaction quotient, Q.
If Q is higher than Ksp, the ions will produce the precipitate CaC₂O₄ until [Ca²⁺][C₂O₄²⁻] = Ksp.
Thus, right answer is:
<em>B. A precipitate will form since Q > Ksp for calcium oxalate</em>
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Answer:
True => ΔH°f for C₆H₆ = 49 Kj/mole
Explanation:
See Thermodynamic Properties Table in appendix of most college level general chemistry texts. The values shown are for the standard heat of formation of substances at 25°C. The Standard Heat of Formation of a substance - by definition - is the amount of heat energy gained or lost on formation of the substance from its basic elements in their standard state. C₆H₆(l) is formed from Carbon and Hydrogen in their basic standard states. All elements in their basic standard states have ΔH°f values equal to zero Kj/mole.
Answer:

Explanation:
Hello!
In this case, since the molarity of magnesium chloride (molar mass = 95.211 g/mol) is 1.672 mol/L and we know the density of the solution, we can first compute the concentration in g/L as shown below:
![[MgCl_2]=1.672\frac{molMgCl_2}{L}*\frac{95.211gMgCl_2}{1molMgCl_2}=159.2\frac{gMgCl_2}{L}](https://tex.z-dn.net/?f=%5BMgCl_2%5D%3D1.672%5Cfrac%7BmolMgCl_2%7D%7BL%7D%2A%5Cfrac%7B95.211gMgCl_2%7D%7B1molMgCl_2%7D%3D159.2%5Cfrac%7BgMgCl_2%7D%7BL%7D)
Next, since the density of the solution is 1.137 g/mL, we can compute the concentration in g/g as shown below:
![[MgCl_2]=159.2\frac{gMgCl_2}{L}*\frac{1L}{1000mL}*\frac{1mL}{1.137g}=0.14](https://tex.z-dn.net/?f=%5BMgCl_2%5D%3D159.2%5Cfrac%7BgMgCl_2%7D%7BL%7D%2A%5Cfrac%7B1L%7D%7B1000mL%7D%2A%5Cfrac%7B1mL%7D%7B1.137g%7D%3D0.14)
Which is also the by-mass fraction and in percent it turns out:

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