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2 years ago

During prophase, which of the following events

Chemistry

1 answer:

alexdok [17]2 years ago

7 0

Answer:

Spindle fibers form.

DNA condenses into chromosomes.

Explanation:

Prophase is the first stage of mitosis. Mitosis produces 2 daughter cells from a parental cell.

Spindle fibers form - this is true. During prophase, the mitotic spindle forms. Later during mitosis the spindle attaches to the centromere of chromosomes and pulls them to opposite ends of the cell prior to division

DNA condenses into chromosomes. - this is true. In the nucleus during interphase, the DNA is relatively loosely compacted. However, prior to division, the DNA is condensed into structures called chromosomes which are then paired up and distributed to the daughter cell.

Chromosomes move to opposite ends of the cell. - this is false. This happens during anaphase

Nuclear membrane begins to re-form. - this is false. This happens during telophase.

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2 years ago

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kramer

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3 years ago

How Many protons is there

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3 years ago

Type in the name of the following compounds. Be sure to use the periodic table if needed.

Snowcat [4.5K]

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2 years ago

Read 2 more answers

Phenolphthalein has a pKa of 9.7 and is colorless in its acid form and pink in its basic form. calculate [In-}/{HIn} for the fol

zvonat [6]

Answer:

1.58x10⁻⁵

2.51x10⁻⁸

0.0126

63.10

Explanation:

Phenolphthalein acts like a weak acid, so in aqueous solution, it has an acid form HIn, and the conjugate base In-, and the pH of it can be calculated by the Handerson-Halsebach equation:

pH = pKa + log[In-]/[HIn]

pKa = -logKa, and Ka is the equilibrium constant of the dissociation of the acid. [X] is the concentrantion of X. Thus,

i) pH = 4.9

4.9 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = - 4.8

[In-]/[HIn] = $10^{-4.8}$

[In-]/[HIn] = 1.58x10⁻⁵

ii) pH = 2.1

2.1 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -7.6

[In-]/[HIn] = $10^{-7.6}$

[In-]/[HIn] = 2.51x10⁻⁸

iii) pH = 7.8

7.8 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -1.9

[In-]/[HIn] = $10^{-1.9}$

[In-]/[HIn] = 0.0126

iv) pH = 11.5

11.5 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = 1.8

[In-]/[HIn] = $10^{1.8}$

[In-]/[HIn] = 63.10

6 0

3 years ago