-dB/dt = kAB = k(2B)(B) = 2kB^2
-dB/B^2 = 2kdt
Integrating: 1/B - 1/(B_0) = 2kt
At t = 10, if 15 g of C have formed, this must have consumed 10 g A and 5 g B. The remaining mass of B is 45 g.
1/45 - 1/50 = 2(k)(10)
k = 1.11 x 10^-4
Then substituting this value of k with t = 40:
1/B - 1/50 = 2(1.11 x 10^-4)(40)
1/B - 1/50 = 0.008889
1/B = 0.028889
B = 34.62 g remaining
Therefore, 50 - 34.62 = 15.38 g of B have been consumed.
Doubling, 30.76 g of A have been consumed.
This means that 15.38 + 30.76 = 46.15 g of C have been formed.
Answer:
True
Explanation:
Carbon dioxide is transported by the blood in the dissolved form. carbonic anhydrase is the enzyme which is a metalloenzyme having zinc at active site converts carbon dioxide into carbonic acid which dissolves in the blood.
Thus,
H₂O (l) + CO₂ (g) ⇔ H⁺(aq) + HCO₃²⁻(aq)
The kidneys in the body help to control the acid base balance. The hydrogen ion secretion in the body leads to the generation and reabsorption of the bicarbonate ions to form carbon dioxide in order to nullify the effect of the acid generated and thus the pH of the blood is maintained.
Apex unfortunately will probably make you get it wrong without adding an "s" into the word break. So if it happens, the answer will be <u>breaks</u>
Answer:
Standard enthalpy for the given reaction is -41.166 kJ
Explanation:
Standard enthalpy of a reaction = ![\sum [n_{i}\times \Delta H_{f}(product)_{i}]-\sum [n_{j}\times \Delta H_{f}(reactant)_{j}]](https://tex.z-dn.net/?f=%5Csum%20%5Bn_%7Bi%7D%5Ctimes%20%5CDelta%20H_%7Bf%7D%28product%29_%7Bi%7D%5D-%5Csum%20%5Bn_%7Bj%7D%5Ctimes%20%5CDelta%20H_%7Bf%7D%28reactant%29_%7Bj%7D%5D)
where 
and 
are number of moles of i-th product and j-th reactant in balanced reaction respectively.
Hence Standard enthalpy for the given reaction = ![[(1\times \Delta H_{f}(CO_{2})_{g})]+[(1\times \Delta H_{f}(H_{2})_{g})]-[(1\times \Delta H_{f}(CO)_{g})]-[(1\times \Delta H_{f}(H_{2}O)_{g})]](https://tex.z-dn.net/?f=%5B%281%5Ctimes%20%5CDelta%20H_%7Bf%7D%28CO_%7B2%7D%29_%7Bg%7D%29%5D%2B%5B%281%5Ctimes%20%5CDelta%20H_%7Bf%7D%28H_%7B2%7D%29_%7Bg%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_%7Bf%7D%28CO%29_%7Bg%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_%7Bf%7D%28H_%7B2%7DO%29_%7Bg%7D%29%5D)
So, Standard enthalpy for the given reaction = ![[(1\times -393.509]+[(1\times 0]-[(1\times -110.525]-[(1\times -241.818]](https://tex.z-dn.net/?f=%5B%281%5Ctimes%20-393.509%5D%2B%5B%281%5Ctimes%200%5D-%5B%281%5Ctimes%20-110.525%5D-%5B%281%5Ctimes%20-241.818%5D)
kJ = -41.166 kJ
