POH=4.22
pOH = -lg[OH⁻]
[OH⁻]=10^(-pOH)
[OH⁻]=10⁻⁴·²² = 6.0×10⁻⁵ mol/L
6.0×10⁻⁵ M
Answer: 253.8
Explanation:
The molar mass of iodine is 126.904. Multiply that by two and you get approximately 253.8 grams in two moles.
Answer: Km = 10μM
Explanation: <u>Michaelis-Menten constant</u> (Km) measures the affinity a enzyme has to its substrate, so it can be known how well an enzyme is suited to the substrate being used. To determine Km another value associated to an eznyme is important: <em>Turnover number (Kcat)</em>, which is the number of time an enzyme site converts substrate into product per unit time.
Enzyme veolcity is calculated as:
![V_{0} = \frac{E_{t}.K_{cat}.[substrate]}{K_{m}+[substrate]}](https://tex.z-dn.net/?f=V_%7B0%7D%20%3D%20%5Cfrac%7BE_%7Bt%7D.K_%7Bcat%7D.%5Bsubstrate%5D%7D%7BK_%7Bm%7D%2B%5Bsubstrate%5D%7D)
where Et is concentration of enzyme catalitic sites and has to have the same unit as velocity of enzyme, so Et = 20nM = 0.02μM;
To calculate Km:
![V_{0}*K_{m} + V_{0}*[substrate] = E_{t}.K_{cat}.[substrate]](https://tex.z-dn.net/?f=V_%7B0%7D%2AK_%7Bm%7D%20%2B%20V_%7B0%7D%2A%5Bsubstrate%5D%20%3D%20E_%7Bt%7D.K_%7Bcat%7D.%5Bsubstrate%5D)
![K_{m} = \frac{E_{t}.K_{cat}.[substrate]-V_{0}*[substrate]}{V_{0}}](https://tex.z-dn.net/?f=K_%7Bm%7D%20%3D%20%5Cfrac%7BE_%7Bt%7D.K_%7Bcat%7D.%5Bsubstrate%5D-V_%7B0%7D%2A%5Bsubstrate%5D%7D%7BV_%7B0%7D%7D)

Km = 10μM
<u>The Michaelis-Menten for the substrate SAD is </u><u>10μM</u><u>.</u>
Answer: Bromine is Actually a Halogen!
Explanation: Bromine is in the 7th column of the Periodic Table. It needs that magic 8 valence electrons and it has 7.
Hoped that helped!