Answer:
Main job of golgi bodies is to sort and package proteins and other substances in a plant cell.
Explanation:
Golgi bodies are also called post office of the cell because it modify and distribute proteins for the cell. First, proteins are made in the organelle of the cell i. e. endoplasmic reticulum. From here, it is send to the Golgi apparatus for modification. Golgi bodies add some special structures with the protein and this protein leaves golgi bodies which is used by the cell where it is needed.
Answer:
the partial of colloidal are smaller and they are not heavy whereas the partial of suspension are larger but they are heavy and less movement thus they settle down due to gravity
Answer:
The answer to your question is: yield = 56.27%
Explanation:
Data
CH3CH2CH2CH2OH (l) → CH3 CH2CH2CH2Br
18.54 ml 1-butanol 15.65 g of 1-bromobutane
% yield = ?
density = 0.81 g/ml
MM = 74 g 1- butanol
MM = 137 g 1-bromobutane
Process
Calculate mass of 1- butanol
density = mass/volume
mass = density x volume
mass = 0.81 x 18.54
mass = 15.02 g of 1-butanol
Theoretical yield
74 g of 1- butanol ----------------- 137 g of 1-bromobutane
15.02 g of 1- butanol ------------- x
x = (15.02 x 137) / 74
x = 27.81 g of 1-bromobutane
% yield = experimental yield / theoretical yield x 100
% yield = 15.65 / 27.81 x 100
% yield = 56.28
<span>Answer is: the mass of hydrogen is 22,05 grams.
m(</span>Al(C₂H₃O₂)₃)<span> = 500 g.
M</span>(Al(C₂H₃O₂)₃) = 27 + 6 ·12 + 9 · 1 + 6 · 16 · g/mol = 204 g/mol.<span>
n</span>(Al(C₂H₃O₂)₃) = m(Al(C₂H₃O₂)₃) ÷ M(Al(C₂H₃O₂)₃).
n(Al(C₂H₃O₂)₃) = 500 g ÷ 204 g/mol.
n(Al(C₂H₃O₂)₃) = 2,45 mol.
n(Al(C₂H₃O₂)₃) : n(H) = 1 : 9.
n(H) = 22,05 mol.
m(H) = 22,05 mol · 1 g/mol
m(H) = 22,05 g.
Answer : The light rays are bent and refraction will occur - B.
