Answer:
4
Step-by-step explanation:
The reactions are:
Glycolysis: 1 glucose ⟶ 2 pyruvate
Link reaction: 2 × [1 pyruvate ⟶ 1 acetyl CoA]
Citric acid cycle: 2 × [1 AcetylCoA ⟶ 2 CO₂]
Now, add the reactions, cancelling species that occur on both sides of the reaction arrow,
1 glucose ⟶ <u>2 pyruvate
</u>
<u>2 pyruvate</u> ⟶ <u>2 acetyl CoA
</u>
<u>2 AcetylCoA </u>⟶ 4 CO₂
<em>Overall</em> : 1 glucose ⟶ 4 CO₂
For each mole of glucose, four molecules of CO₂ are released in the citric acid cycle.
Answer:

Explanation:
Hello!
In this case for the solution you are given, we first use the mass to compute the moles of CuNO3:

Next, knowing that the molarity has units of moles over liters, we can solve for volume as follows:

By plugging in the moles and molarity, we obtain:

Which in mL is:

Best regards!
The theobromine molecule contains a total of 22 bonds
The question is incomplete, the complete question is;
Chlorine monoxide accumulates in the stratosphere above Antarctica each winter 3nd plays a key role the formation of the ozone hole above the South Pole each spring Eventually. CIO decomposes acco to the equation: 2CIO(g) rightarrow CL2(g) + O2(g) The second-order rate constant for the decomposition of CIO is 6950000000 M-1 s-1 at a particular temperature Determine the half-life of CIO when its initial concentration is .0000000185 M
Answer:
7.8 * 10^-3 s
Explanation:
Given that the half life of a second order reaction is obtained from the formula;
t1/2 = k-1[A]o-1
t1/2 = 1/k[A]o
second order rate constant (k) = 6950000000 M-1 s-1
initial concentration ([A]o) =0.0000000185 M
t1/2 = 1/6950000000 * 0.0000000185
t1/2 = 7.8 * 10^-3 s