Answer:- 0.273 kg
Solution:- A double replacement reaction takes place. The balanced equation is:
We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.
= 63.8 g aluminum nitrate
From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.
We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.
= 
sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate and then the grams are converted to kg.
= 0.273 kg
So, 0.273 kg of 35% m/m sodium chlorate solution are required.
Answer:
Percentage abundance of 121 Sb is = 57.2 %
Percentage abundance of 123 Sb is = 42.8 %
Explanation:
The formula for the calculation of the average atomic mass is:
Given that:
Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.
For first isotope, 121 Sb :
% = x %
Mass = 120.9038 u
For second isotope, 123 Sb:
% = 100 - x
Mass = 122.9042 u
Given, Average Mass = 121.7601 u
Thus,
Solving for x, we get that:
x = 57.2 %
<u>Thus, percentage abundance of 121 Sb is = 57.2 %
</u>
<u>percentage abundance of 123 Sb is = 100 - 57.2 % = 42.8 %</u>
Answer:
Explanation:
A protein is a long chain of amino acids linked together by amide groups.
The general structure is
![\rm \left[-NHCHR-\underbrace{\hbox{CO-NH}}_{\hbox{amide group}}-CHRCO-\right]_{n}](https://tex.z-dn.net/?f=%5Crm%20%5Cleft%5B-NHCHR-%5Cunderbrace%7B%5Chbox%7BCO-NH%7D%7D_%7B%5Chbox%7Bamide%20group%7D%7D-CHRCO-%5Cright%5D_%7Bn%7D)
Im not 100% but i think b
Answer:
Here's what I find.
Explanation:
a. Structure
Acebutolol is a secondary amine (basic). It forms a substituted ammonium salt when treated with hydrochloric acid.
The structure of the salt is shown below, with a red arrow pointing toward the positive charge on the N atom.
b. Solubility
The formula of acebutolol is C₁₈H₂₈N₂O₄.
The amide, acetyl, and ether groups confer little solubility to the molecule.
The alcohol and secondary amine do confer some solubility, because they can donate and accept hydrogen bonds.
However, they can each overcome the hydrophobic properties of only three to five carbons, and acebutolol has 18 of them.
The free amine would be preferentially soluble in lipid material (fats)
The protonated amine is ionic and therefore much more soluble in aqueous media (e.g., blood).
c. Marketing
The drug must be delivered to the tissues of the heart, where it blocks the effects of adrenalin. The best way to do this is through the blood, so acebutolol is marketed as the hydrochloride salt.
