Answer:
a) carboxylic acid
b) amine
c) ester
d) aldehyde
e) alkene
f) ketone
Explanation:
Most of them are straight forward.
It's found inside of a cell attached to the cytoplasm and attached to the endoplasmic reticulum. hope this helps
Explanation:
The given data is as follows.
Solvent 1 = benzene, Solvent 2 = water
= 2.7, 
= 100 mL
= 10 mL, weight of compound = 1 g
Extract = 3
Therefore, calculate the fraction remaining as follows.
![f_{n} = [1 + K_{p}(\frac{V_{S_{2}}}{V_{S_{1}}})]^{-n}](https://tex.z-dn.net/?f=f_%7Bn%7D%20%3D%20%5B1%20%2B%20K_%7Bp%7D%28%5Cfrac%7BV_%7BS_%7B2%7D%7D%7D%7BV_%7BS_%7B1%7D%7D%7D%29%5D%5E%7B-n%7D)
= ![[1 + 2.7(\frac{100}{10})]^{-3}](https://tex.z-dn.net/?f=%5B1%20%2B%202.7%28%5Cfrac%7B100%7D%7B10%7D%29%5D%5E%7B-3%7D)
= 
= 
Hence, weight of compound to be extracted = weight of compound - fraction remaining
= 1 -
= 0.00001
or, = 
Thus, we can conclude that weight of compound that could be extracted is .
Mg2+...because it loses 2 electron which is on its valence shell making it empty and making it have jst 2 shells. The radius calculated from d nucleus to the new valence shell is then smaller than that of Cl^- and k+
Answer:
The equation: (NH₄)₂SO₄ = 2NH4(+) + SO4(-2)
The number of moles = 5 g / 132.14 g/mol = 0.038 mol
The number of molecules = 0.038 X 6.022x10^23 = 2.29x10^23
the number of positive ions present in the ammonium sulphate solution:
2 positive ions for every 1 molecule of (NH₄)₂SO₄
so 2 x 2.29x10^23 = 4.58x10^23
the number of negative ions present in the ammonium sulphate solution
1 negative ion for every 1 molecule of (NH₄)₂SO₄
so 1 x 2.29x10^23 = 2.29x10^23
the total number of ions present in the ammonium sulphate solution
4.58x10^23 + 2.29x10^23 = 6.87x10^23
