Answer:
the food kingdom
Explanation:
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The four ionic species initially in solution are Na⁺, PO₄³⁻, Cr³⁺, and Cl⁻. Since the precipitate is composed of Cr³⁺ and PO₄³⁻ ions, the spectator ions must be Na⁺ and Cl⁻.
The complete ionic equation is 3Na⁺(aq) + PO₄³⁻(aq) + Cr₃⁺(aq) + 3Cl⁻(aq) → 3Na⁺(aq) + 3Cl⁻(aq) + CrPO₄(s).
So the balanced <u>net ionic equation</u> for this reaction would be Cr³⁺(aq) + PO₄³⁻(aq) → CrPO₄(s).
Answer:
= 19
ΔG° of the reaction forming glucose 6-phosphate = -7295.06 J
ΔG° of the reaction under cellular conditions = 10817.46 J
Explanation:
Glucose 1-phosphate ⇄ Glucose 6-phosphate
Given that: at equilibrium, 95% glucose 6-phospate is present, that implies that we 5% for glucose 1-phosphate
So, the equilibrium constant
can be calculated as:
![= \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%5Bglucose-6-phosphate%5D%7D%7B%5Bglucose-1-%5Bphosphate%5D%7D)


= 19
The formula for calculating ΔG° is shown below as:
ΔG° = - RTinK
ΔG° = - (8.314 Jmol⁻¹ k⁻¹ × 298 k × 1n(19))
ΔG° = 7295.05957 J
ΔG°≅ - 7295.06 J
b)
Given that; the concentration for glucose 1-phosphate = 1.090 x 10⁻² M
the concentration of glucose 6-phosphate is 1.395 x 10⁻⁴ M
Equilibrium constant
can be calculated as:
![= \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%5Bglucose-6-phosphate%5D%7D%7B%5Bglucose-1-%5Bphosphate%5D%7D)

0.01279816514 M
0.0127 M
ΔG° = - RTinK
ΔG° = -(8.314*298*In(0.0127)
ΔG° = 10817.45913 J
ΔG° = 10817.46 J
Answer:
Pb(C2H3O2)2 + KI ----> PbI2 + KC2H3O2
Explanation:
all the numbers are written as subscripts
To determine the amplitude of a transverse wave, measure the highest amount of disturbance from the equilibrium that the wave experiences.