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2 years ago

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The cell cycle can be divided into two phases interphase and mitosis (cell division), Mitosis is further subdivided into prophas

e, metaphase, anaphase, and telophase.
A lab technician observed 200 cells from a cell line and recorded the number of cells in each phase of the cell cycle. The results are shown in the table.
Number of Cells Observed
(out of 200)
Phase Number of Cells Observed
Interphase
170
prophase
metaphase
anaphase
telophase
Based on the data, what is the probability a randomly chosen cell will be observed undergoing cell division?
You may use the calculator,
A 15%
B. 18%
C 30%
D. 85%

2 answers:

stealth61 [152]2 years ago

7 0

Well the answer is C

docker41 [41]2 years ago

3 0

the answer to this is c i hope this helped

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_________is a process in which o2 is released as a by-product of oxidation-reduction reactions

Kisachek [45]

Answer : The combustion is a process in which oxygen is released as a by-product of oxidation-reduction reactions.

Explanation :

Combustion reaction : It is defined as the reactions in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide and water.

The chemical equation of combustion reaction is:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

The combustion reaction is also a redox reaction.

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

The combustion reaction is also a redox reaction in which the carbon shows oxidation by the addition of oxygen or removal of hydrogen and oxygen shows reduction by the addition of hydrogen or removal of oxygen.

Hence, the combustion is a process in which oxygen is released as a by-product of oxidation-reduction reactions.

6 0

2 years ago

A bottle of concentrated aqueous sulfuric acid, labeled 98.0 wt% h2so4, has a concentration of 18.0 m. (a) how many milliliters

nadya68 [22]

<span>n this order, Ď=1.8gmL, cm=0.5, and mole fraction = 0.9 First, let's start with wt%, which is the symbol for weight percent. 98wt% means that for every 100g of solution, 98g represent sulphuric acid, H2SO4. We know that 1dm3=1L, so H2SO4's molarity is C=nV=18.0moles1.0L=18M In order to determine sulphuric acid solution's density, we need to find its mass; H2SO4's molar mass is 98.0gmol, so 18.0moles1Lâ‹…98.0g1mole=1764g1L Since we've determined that we have 1764g of H2SO4 in 1L, we'll use the wt% to determine the mass of the solution 98.0wt%=98g.H2SO4100.0g.solution=1764gmasssolutionâ†’ masssolution=1764gâ‹…100.0g98g=1800g Therefore, 1L of 98wt% H2SO4 solution will have a density of Ď=mV=1800g1.0â‹…103mL=1.8gmL H2SO4's molality, which is defined as the number of moles of solute divided by the mass in kg of the solvent; assuming the solvent is water, this will turn out to be cm=nH2SO4masssolvent=18moles(1800â’1764)â‹…10â’3kg=0.5m Since mole fraction is defined as the number of moles of one substance divided by the total number of moles in the solution, and knowing the water's molar mass is 18gmol, we could determine that 100g.solutionâ‹…98g100gâ‹…1mole98g=1 mole H2SO4 100g.solutionâ‹…(100â’98)g100gâ‹…1mole18g=0.11 moles H2O So, H2SO4's mole fraction is molefractionH2SO4=11+0.11=0.9</span>

5 0

2 years ago

A 40.0 mL sample of 0.25 M KOH is added to 60.0 mL of 0.15 M Ba(OH)2. What is the molar concentration of OH-(aq) in the resultin

solmaris [256]

Answer:

C) 0.28 M

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Potassium hydroxide will furnish hydroxide ions as:

$KOH\rightarrow K^{+}+OH^-$

Given :

<u>For Potassium hydroxide : </u>

Molarity = 0.25 M

Volume = 40.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 40.0×10⁻³ L

Thus, moles of hydroxide ions furnished by Potassium hydroxide is same as the moles of Potassium hydroxide as shown below:

$Moles =0.25 \times {40.0\times 10^{-3}}\ moles$

Moles of hydroxide ions by Potassium hydroxide = 0.01 moles

Barium hydroxide will furnish hydroxide ions as:

$Ba(OH)_2\rightarrow Ba^{2+}+2OH^-$

Given :

<u>For Barium hydroxide : </u>

Molarity = 0.15 M

Volume = 60.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 60.0×10⁻³ L

Thus, moles of hydroxide ions furnished by Barium hydroxide is twice the moles of Barium hydroxide as shown below:

$Moles =2\times 0.15 \times {60.0\times 10^{-3}}\ moles$

Moles of hydroxide ions by Barium hydroxide = 0.018 moles

Total moles = 0.01 moles + 0.018 moles = 0.028 moles

Total volume = 40.0×10⁻³ L + 60.0×10⁻³ L = 100×10⁻³ L

Concentration of hydroxide ions is:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity_{OH^-}=\frac{0.028 }{100\times 10^{-3}}$

<u> The final concentration of hydroxide ion = 0.28 M</u>

5 0

2 years ago

Read 2 more answers

Consider the following reactions and their respective equilibrium constants: NO(g)+12Br2(g)⇌NOBr(g)Kp=5.3 2NO(g)⇌N2(g)+O2(g)Kp=2

maxonik [38]

Answer:

Equilibrium constant of the given reaction is $1.3\times 10^{-29}$

Explanation:

$NO+\frac{1}{2}Br_{2}\rightleftharpoons NOBr$ .... $(K_{p})_{1}=5.3$

$2NO\rightleftharpoons N_{2}+O_{2}$ .... $(K_{p})_{2}=2.1\times 10^{30}$

The given reaction can be written as summation of the following reaction-

$2NO+Br_{2}\rightleftharpoons 2NOBr$

$N_{2}+O_{2}\rightleftharpoons 2NO$

......................................................................................

$N_{2}+O_{2}+Br_{2}\rightleftharpoons 2NOBr$

Equilibrium constant of this reaction is given as-

$\frac{[NOBr]^{2}}{[N_{2}][O_{2}][Br_{2}]}$

$=(\frac{[NOBr]}{[NO][Br_{2}]^{\frac{1}{2}}})^{2}(\frac{[NO]^{2}}{[N_{2}][O_{2}]})$

$=\frac{(K_{p})_{1}^{2}}{(K_{p})_{2}}$

$=\frac{(5.3)^{2}}{2.1\times 10^{30}}=1.3\times 10^{-29}$

7 0

2 years ago

Read 2 more answers

Which of the following type of protons are chemically equivalent? A) homotopic B) enantiotopic C) diastereotopic A & B B &am

Rina8888 [55]

Answer:

A) homotopic and B) enantiotopic

Explanation:

Protons chemically equivalent are those that have the same chemical shift, also if they are interchangeable by some symmetry operation or by a rapid chemical process.

The existence of symmetry axes, Cn, that relate to the protons results in the protons being homotopic, that is chemically equivalent in both chiral and aquiral environments.

The existence of a plane of symmetry, σ, makes the protons related by it, are enantiotopic and these protons will only be equivalent in an aquiral medium; if the medium is chiral both protons will be chemically NOT equivalent. The existence of a center of symmetry, i, in the molecule makes the related protons through it enantiotopic and therefore chemically only in the aquiral medium.

Diastereotopic protons cannot be interconverted by any symmetry operation and they are different, with different chemical displacement.

6 0

2 years ago