Answer:
Stretching is the lengthening and contraction is the shortening
Explanation:
Unless it's asking for eccentric as the stretching and concentric as the shortening
Answer:
Is there any other part to this question? If not I'm pretty sure the answer is 205.5 kJ
Explanation:
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:
just see it it will help trust me its my school work
Explanation:
1. Find its coordination figure/coordination number of central atom (CF)
Ev = Vallence electron of central atom
Σe = electrons donated from substituents
Terminal O gives 0 electrons, hence Σe = 3 x 0
charge = charge of the compound
2. Find EP (electron pairs) and LP (lone pairs)
LP = CF - EP
3. Draw the skeleton with octet substituents (top right figure)
4. Find formal charge for each atoms (Qf)
5. Write formal charge near atom in skeleton
6. Enjoy
