I believe the answer is 4 carbons. Glycolysis involves break down of glucose to two molecules of pyruvic acid (3 carbons) under aerobic conditions. At the end of glycolysis the two pyruvate molecules undergoes pyruvate oxidation to capture the remaining energy in the form of ATP. A carboxyl group is removed from pyruvate and released in the form carbon dioxide, leaving a two carbon molecule which forms Acetyl-CoA (2 molecules). Acetyl-CoA then serves as a fuel for the citric acid cycle in the next stage of cellular respiration.
Explanation:
Ksp of NiCO3 = 1.4 x 10^-7
Ksp of CuCO3 = 2.5 x 10^-10
Ionic equations:
NiCO3 --> Ni2+ + CO3^2-
CuCO3 --> Cu2+ + CO3^2-
[Cu2+][CO3^2-]/[Ni2+][CO3^2-]
= (2.5* 10^-10)/(1.4* 10^-7)
= 0.00179.
[Cu2+]/[Ni2+]
= 0.00179
= 0.00179*[Ni2+]
If all of Cu2+ is precipitated before Na2CO3 is added.
= 0.00179 * (0.25)
The amount of Cu2+ not precipitated = 0.000448 M
The percent of Cu2+ precipitated before the NiCO3 precipitates = concentration of Cu2+ unprecipitated/initial concentration of Cu2+ * 100
= 0.000448/0.25 * 100
= 0.18%
Therefore, percentage precipitated = 100 - 0.18
= 99.8%
The two metal ions can be separated by slowly adding Na2CO3. Thus that is the unpptd Cu2+.
I don’t really know this but I think
Answer:
The concentration of COF₂ at equilibrium is 0.296 M.
Explanation:
To solve this equilibrium problem we use an ICE Table. In this table, we recognize 3 stages: Initial(I), Change(C) and Equilibrium(E). In each row we record the <em>concentrations</em> or <em>changes in concentration</em> in that stage. For this reaction:
2 COF₂(g) ⇌ CO₂(g) + CF₄(g)
I 2.00 0 0
C -2x +x +x
E 2.00 - 2x x x
Then, we replace these equilibrium concentrations in the Kc expression, and solve for "x".
![Kc=8.30=\frac{[CO_{2}] \times [CF_{4}] }{[COF_{2}]^{2} } =\frac{x^{2} }{(2.00-2x)^{2} } \\8.30=(\frac{x}{2.00-2x} )^{2} \\\sqrt{8.30} =\frac{x}{2.00-2x}\\5.76-5.76x=x\\x=0.852](https://tex.z-dn.net/?f=Kc%3D8.30%3D%5Cfrac%7B%5BCO_%7B2%7D%5D%20%5Ctimes%20%5BCF_%7B4%7D%5D%20%7D%7B%5BCOF_%7B2%7D%5D%5E%7B2%7D%20%7D%20%3D%5Cfrac%7Bx%5E%7B2%7D%20%7D%7B%282.00-2x%29%5E%7B2%7D%20%7D%20%5C%5C8.30%3D%28%5Cfrac%7Bx%7D%7B2.00-2x%7D%20%29%5E%7B2%7D%20%5C%5C%5Csqrt%7B8.30%7D%20%3D%5Cfrac%7Bx%7D%7B2.00-2x%7D%5C%5C5.76-5.76x%3Dx%5C%5Cx%3D0.852)
The concentration of COF₂ at equilibrium is 2.00 -2x = 2.00 - 2 × 0.852 = 0.296 M
Explanation:
number of moles = number of particle/ 6.02X10^23
= 12.044 X 10^24/6.02×10^23
=20.0066 moles so round it up to 3 S.F
So
=20.0moles
