Mitosis is conventionally divided into 5 phases, which include prophase, prometaphase, metaphase, anaphase and telophase and cytokinesis.
Interphase
Before coming into mitosis, a mobile spends a length of its increase underneath interphase.
Prophase
Prophase straight away follows the S and G2 levels of the cycle and is marked by way of condensation of the genetic fabric to form compact mitotic chromosomes composed of chromatids attached at the centromere.
Prometaphase
In the prometaphase, the nuclear envelop disintegrates. Now the microtubules are allowed to extend from the centromere to the chromosome.
Metaphase
At this level, the microtubules start pulling the chromosomes with equal pressure and the chromosome ends up in the center of the cell. This area is referred to as the metaphase plate.
Anaphase
The splitting of the sister chromatids marks the onset of anaphase. These sister chromatids end up the chromosome of the daughter nuclei.
Telophase
The chromosomes that cluster at the two poles start coalescing into an undifferentiated mass, because the nuclear envelope begins forming round it.
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Answer:
Total energy required to raise the temperature of 425 g of tin from 298.15 K to 505.05 K and to melt the tin at 505.05 K is 45.249 kiloJoules.
Explanation:
Mass of the tin ,m= 425 g
Heat capacity of the tin ,c= 0.227 J/g K
Initial temperature of the tin ,
= 25.0 °C = 298.15 K
Final temperature of the tin, 
= 231.9 °C = 505.05 K
Let the heat required to change the temperature of tin from 298.15 K to 505.05 K be Q.
Heat required to melt tin at 505.05 K be Q'
The heat of fusion of tin metal =
Total energy required to raise the temperature of 425 g of tin from 298.15 K to 505.05 K and to melt the tin at 505.05 K is:
= Q+Q' = 19.961 kJ + 25.288 kJ = 45.249 kJ
Answer:
Explanation:
The amount adsorbed (solute) is the acetic acid, and the adsorbent is the activated charcoal. The mass of the adsorbent is 10 g.
So, we need to calculate the mass of the acetic acid as follows:
Where:
n: is the number of moles = C*V
M: is the molecular mass = 60.052 g/mol
C: is the final concentration of the acid = 0.5*0.2 mol/L = 0.10 mol/L
V: is the volume = 50 ml = 0.050 L
Now, the amount of solute adsorbed per gram of the adsorbent is:
Therefore, the amount of solute adsorbed per gram of the adsorbent is 0.03 g/g.
I hope it helps you!
Answer:
Option D
Explanation:
Displacement is a vector quantity.
Answer:
False
Explanation:
using the measurements of the amounts of chemicals, it can be calculated to some degree of accuracy depending mainly on the accuracy of the measurements and enviroments of the reaction
