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.
Answer:
i hope my answer help u :))
Explanation:
no , it will be 39.99711
Answer:
6.82 g H₂S
General Formulas and Concepts:
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
0.200 mol H₂S
<u>Step 2: Identify Conversions</u>
Molar Mass of H - 1.01 g/mol
Molar Mass of S - 32.07 g/mol
Molar Mass of H₂S - 2(1.01) + 32.07 = 34.09 g/mol
<u>Step 3: Convert</u>
- Set up:
- Multiply:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
6.818 g H₂S ≈ 6.82 g H₂S
Answer:
Fe3+ + 3e– --------> Fe
Explanation:
Reduction refers to a gain of electrons. A specie is said to be reduced when it gains electrons or when it experiences a decrease in oxidation number. The both sentences above can be clearly seen when inspecting a reduction half equation.
Consider the redox reaction half equation;
Fe3+ + 3e– ------>Fe
We can see that Fe^3+ accepted three electrons, the oxidation number of iron thereby decreased from +3 to zero from left to right. This implies that the Fe^3+ was actually reduced according to the equation shown.
Empirical=12.01+2.02+16=30.03
Molecular=60
Molecular/Empirical=2
so, 2(CH2O)
C2H4O2