Answer:
964ug
Explanation:
The problem here involves converting from one unit to another.
We are to convert from ounces to micrograms.
1ug = 1 x 10⁻⁶g
1oz = 28.35g
So we first convert to grams from oz then take to ug:
Solving:
1oz = 28.35g
3.4 x 10⁻⁵oz will then give 3.4 x 10⁻⁵ x 28.35 = 9.64 x 10⁻⁴g
So;
1 x 10⁻⁶g = 1ug
9.64 x 10⁻⁴g will give 
= 9.64 x 10²ug or 964ug
Answer:
Tamoxifen is an irreversible, competitive inhibitor.
Explanation:
In order to binds to the active site of the estrogen receptor protein, tamoxifen have to compete with the other chemical compound, and inhibits the estrogen release, so it is a competitive inhibitor. Then, you said that when tamoxifen binds to the receptor, the protein is permanently deactivated, so it is also irreversible.
Answer:
pH 9,8 is likely to work best for this separation
Explanation:
Ion exchange chromatography is a chemical process where molecules are separated by affinity to an ion exchange resin. To separate different aminoacids you must use the isoelectric point (That is the pH where the aminoacid will be in its neutral form).
For lysine, PI is:
9,8
For arginine:
10,75
At pH = 9,8 lysine will be in its neutral form and will not be retain in the column but arginine will be in +1 charge being retained by the ion exchange resin.
Thus, <em>pH 9,8 is likely to work best for this separation</em>
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I hope it helps!
During the dark ages-All that wasn't around the christian religion were lost. Knowledge and discoveries before the Dark Ages were mainly in Greek ideas. Biblic scripts however were in Latin. Therefore all inventions up to the dark ages were lost and only religious scripts preserved in the churches. Outside the church, no one knew how to write, read or even communicate.
Answer:
a) pH = 4.213
b) % dis = 2 %
Explanation:
Ch3COONa → CH3COO- + Na+
CH3COOH ↔ CH3COO- + H3O+
∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]
mass balance:
⇒ <em>C</em> CH3COOH + <em>C</em> CH3COONa = [ CH3COOH ] + [ CH3COO- ]
<em>∴ C </em>CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)
∴ <em>C</em> CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL
⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]
charge balance:
⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water
⇒ [ CH3COO- ] = [ H3O+ ] + 1.00
⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5
⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]
⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0
⇒ [ H3O+ ] = 6.12 E-5 M
⇒ pH = - Log [ H3O+ ] = 4.213
b) (% dis)* mol acid = <em>C</em> CH3COOH = 3.4
∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol
⇒ % dis = 3.4 / 1.7 = 2 %
