Answer:
Sun and the stars, nothing humans don't use fusion at all
Explanation:
Answer:
c. rate=−1/2Δ[HBr]/Δt=Δ[H2]/Δt=Δ[Br2]/Δt
Explanation:
Hello,
In this case, the undergoing chemical reaction is:

Thus, the rate is given as:
![rate=-\frac{1}{2} \frac{\Delta [HBr]}{\Delta t}=\frac{\Delta [Br_2]}{\Delta t} =\frac{\Delta [H_2]}{\Delta t}](https://tex.z-dn.net/?f=rate%3D-%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7B%5CDelta%20%5BHBr%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B%5CDelta%20%5BBr_2%5D%7D%7B%5CDelta%20t%7D%20%3D%5Cfrac%7B%5CDelta%20%5BH_2%5D%7D%7B%5CDelta%20t%7D)
It is necessary to remember that each concentration to time interval is divided into the stoichiometric coefficient, that is why HBr has a 1/2. Moreover, the concentration HBr is negative since it is a reactant and it has a negative rate due to its consumption.
Therefore, the answer is:
c. rate=−1/2Δ[HBr]/Δt=Δ[H2]/Δt=Δ[Br2]/Δt
Best regards.
Answer
147.06N
Explanation:
MA=Load/Effort
6.8=1000/x
Finding x you make x the subject and divide 1000 by 6.8 which is 147.06N
Answer:
The ionization equation is
⇄
(1)
Explanation:
The ionization equation is
⇄
(1)
As the Bronsted definition sais, an acid is a substance with the ability to give protons thus, H2PO4 is the acid and HPO42- is the conjugate base.
The Ka expression is the ratio between the concentration of products and reactants of the equilibrium reaction so,
![Ka = \frac{[HPO_{4}^{-2}] [H_{3}O^{+}]}{[H_{2}PO_{4}^{-}] [H_{2}O]} = 6.2x10^{-8}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BHPO_%7B4%7D%5E%7B-2%7D%5D%20%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5BH_%7B2%7DPO_%7B4%7D%5E%7B-%7D%5D%20%5BH_%7B2%7DO%5D%7D%20%3D%206.2x10%5E%7B-8%7D)
The pKa is

The pKa of H2CO3 is 6,35, thus this a stronger acid than H2PO4. The higher the pKa of an acid greater the capacity to donate protons.
In the body H2CO3 is a more optimal buffer for regulating pH due to the combination of the two acid-base equilibriums and the two pKa.
If the urine is acidified, according to Le Chatlier's Principle the equilibrium (1) moves to the left neutralizing the excess proton concentration.