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.
Answer:
20 atoms
Explanation:
There are 4 in H2O2 because of 2 hydrogens and 2 oxygens.
Then, multiply by 5 because the coefficient is 5, therefore there are 5 H2O2 molecules.
5 x 4 = 20
Answer:
3.96 MINUTES
Explanation:
Karl likes to use cruise control when he drives on
the highway. After setting cruise control at 33.53
m/s (75 mph). At this speed, how long would it
take Karl to travel 8 kilometers?
VELOCITY = DISTANCE/TIME
OR
V=D/T
SO
VT=D
SO
T=D/V
THE DISTANCE IS 8 KM
THE VELOCITY IS 75 MPH
THERE ARE 1.61 KM PER MILE
SO
75M[H =
75 X 1.61 KM/H =
121 KM/HR
T=D/V
T=8/121 =0.066 HOUR =
0.066 X 60 MINUTES =
3.96 MINUTES
CHECK
75 MPH =
(75 X 1.61) KM/60 MINUTES =
120,75 KM/60 MINUTES=
2.01 KM/MINUTE
to travel 8 K,M WILL TAKE
8/2.01 = 3.98 minutes
If one or more of the following occur: Gas formation, precipitate formation, color change, temperature change, or odor
Answer:
Option 4 ) 1-butyne
Explanation:
In organic chemistry, you should use IUPAC convention in order to name an organic compound. First of all, you should identify the lenght of the organic chain, for this case, you have 5 C atoms, but in fact, you have a triple bond (which would be a substitute: ethynil-) as a substitute, so the main skeleton would have 4 C atoms (a butane)
Then, you start by numbering carbon N° 1 as the one that has the substitute (triple bound)-starting from the right, it would be the second C):
CH₃-CH₂-CH₂-C≡CH
Which will finally leads us to 1-butyne
