
- <em>Expansion </em><em>of </em><em>particles</em><em> </em><em>of</em><em> </em><em>substances.</em><em> </em>
- <em>Increase</em><em> </em><em>in </em><em>temperature</em><em>.</em>
- <em>Change</em><em> </em><em>in </em><em>state</em><em>.</em>
- <em>Change</em><em> </em><em>in </em><em>physical</em><em> </em><em>property</em>
- <em>It </em><em>may </em><em>bring</em><em> </em><em>out </em><em>chemical</em><em> </em><em>changes</em><em>.</em>
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Molecules arranged in regular pattern changes into an irregular pattern
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>
<em></em>
Answer:
A) SiO2 is the limiting reactant
B) Theoretical yield= 72333.3g
C) % yield =91.5%
Explanation:
SiO2(s) + 2C(s) --------------> Si(s) + 2CO(g)
n(SiO2)= 155000/60 = 2583.33 mols
n(C)= 79000/12= 3291.66 mols
a)SiO2 is the limiting reactant
According to the balanced reaction equation,
60g of SiO2 produced 28g of SiO2
155000g of SiO2 will produce 155000×28/60= 72333.3g
Therefore theoretical yield of Si= 72333.3g
% yield= 66200/72333.3×100/1 =91.5%