The answer is False. the amplitude shows how high or low something is
The correct option is A.
To calculate the binding energy, you have to find the mass defect first.
Mass defect = [mass of proton and neutron] - Mass of the nucleus
The molar mass of thorium that we are given in the question is 234, the atomic number of thorium is 90, that means the number of neutrons in thorium is
234 - 90 = 144.
The of proton in thourium is 90, same as the atomic number.
Mass defect = {[90 * 1.00728] +[144* 1.00867]} - 234
Note that each proton has a mass of 1.00728 amu and each neutron has the mass of 1.00867 amu.
Mass defect = [90.6552 + 145.24848] - 234 = 1.90368 amu.
Note that the unit of the mass is in amu, it has to be converted to kg
To calculate the mass in kg
Mass [kg] = 1.90368 * [1kg/6.02214 * 10^-26 = 3.161135 * 10^-27
To calculate the binding energy
E = MC^2
C = Speed of light constant = 2.9979245 *10^8 m/s2
E = [3.161135 * 10^-27] * [2.9979245 *10^8]^2
E = 2.84108682069 * 10^-10.
Note that we arrive at this answer because of the number of significant figures that we used.
So, from the option given, Option A is the nearest to the calculated value and is our answer for this problem.
What were Lamarck's ideas about evolution and why were those ideas incorrect
To get the concentration of the second solution let us use the following formulae
C1V1=C2V2 where C1 is concentration of first solution and V1 is the volume of solution first solution. on the other hand C2 is the concentration of second solution and V2 is the volume of second solution.
therefore
0.8×2=(2+10)×C2
1.6 =12×C2
1.6/12=C2
C2 = 0.1333mg/mL
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.
To know more about mitosis at
brainly.com/question/8757261
