Answer: I don't know if this helps but here's some information I think might help.
Usually, cells will take between 5 and 6 hours to complete S phase. G2 is shorter, lasting only 3 to 4 hours in most cells. In sum, then, interphase generally takes between 18 and 20 hours. Mitosis, during which the cell makes preparations for and completes cell division only takes about 2 hours. Calculate the percentage of time spent in each phase by counting the total number of cells in each phase (total in interphase, in prophase, etc.) and dividing each by the total number of cells you counted. How do cancer cells differ in total time required for mitosis? Cancer cells produce 117 minutes faster than regular cells. ... Normal cells require 640 minutes during interphase, cancer cells only need 380. For prophase, cancerous cells need 15 minutes less than regular cells. Another hallmark of cancer cells is their "replicative immortality," a fancy term for the fact that they can divide many more times than a normal cell of the body. In general, human cells can go through only about 40-60 rounds of division before they lose the capacity to divide, "grow old," and eventually die 3.
B) bacteria
generally less developed organism perform asexual method
Answer : a) 40075.14 J/mol
Explanation :
According to the Arrhenius equation,
or,
where,
= rate constant at = k
= rate constant at = 10 k
= activation energy for the reaction = ?
R = gas constant = 8.314 J/mole.K
= initial temperature = 440 K
= final temperature = 550 K
Now put all the given values in this formula, we get
:
Therefore, the activation energy for the reaction is 40075.14J/mol.
Answer:
<h2>The first thing to do here is to use the molarity and the volume of the initial solution to figure out how many grams of copper(II) chloride it contains.</h2><h2 /><h2>133</h2><h2>mL solution</h2><h2>⋅</h2><h2>1</h2><h2>L</h2><h2>10</h2><h2>3</h2><h2>mL</h2><h2>⋅</h2><h2>7.90 moles CuCl</h2><h2>2</h2><h2>1</h2><h2>L solution</h2><h2>=</h2><h2>1.051 moles CuCl</h2><h2>2</h2><h2 /><h2>To convert this to grams, use the compound's molar mass</h2><h2 /><h2>1.051</h2><h2>moles CuCl</h2><h2>2</h2><h2>⋅</h2><h2>134.45 g</h2><h2>1</h2><h2>mole CuCl</h2><h2>2</h2><h2>=</h2><h2>141.31 g CuCl</h2><h2>2</h2><h2 /><h2>Now, you know that the diluted solution must contain </h2><h2>4.49 g</h2><h2> of copper(II) chloride. As you know, when you dilute a solution, you increase the amount of solvent while keeping the amount of solute constant.</h2><h2 /><h2>This means that you must figure out what volume of the initial solution will contain </h2><h2>4.49 g</h2><h2> of copper(II) chloride, the solute.</h2><h2 /><h2>4.49</h2><h2>g</h2><h2>⋅</h2><h2>133 mL solution</h2><h2>141.32</h2><h2>g</h2><h2>=</h2><h2>4.23 mL solution</h2><h2>−−−−−−−−−−−−−− </h2><h2 /><h2>The answer is rounded to three sig figs.</h2><h2 /><h2>You can thus say that when you dilute </h2><h2>4.23 mL</h2><h2> of </h2><h2>7.90 M</h2><h2> copper(II) chloride solution to a total volume of </h2><h2>51.5 mL</h2><h2> , you will have a solution that contains </h2><h2>4.49 g</h2><h2> of copper(II) chloride.</h2>