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3 years ago

_______is the division of the cytoplasm at the end of mitosis.

Chemistry

1 answer:

tankabanditka [31]3 years ago

6 0

Answer: D. CYTOKINESIS

Explanation:

Mitosis<em> </em>ends<em> with telophase, or the stage at which the chromosomes reach the poles. ... Telophase is followed by </em>CYTOKINESIS<em>, or the division of the cytoplasm into two daughter </em><em>cells.</em>

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2 years ago

while doing an experiment you got stuck and didn't know how to proceed what will you do to continue with the experiment

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Answer:

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3 years ago

Before tackling this problem, be sure you know how to find the antilog of a number using a scientific calculator.

dybincka [34]

<h2>Question:- </h2>

A solution has a pH of 5.4, the determination of [H+].

<h2>Given :- </h2>

pH:- 5.4
pH = - log[H+]

<h2>To find :- concentration of H+</h2>

<h2>Answer:- Antilog(-5.4) or 4× 10-⁶</h2>

<h2>Explanation:- </h2><h3>Formula:- pH = -log H+ </h3>

Take negative to other side

-pH = log H+

multiple Antilog on both side

(Antilog and log cancel each other )

Antilog (-pH) = [ H+ ]

New Formula :- Antilog (-pH) = [+H]

Now put the values of pH in new formula

Antilog (-5.4) = [+H]

we can write -5.4 as (-6+0.6) just to solve Antilog

Antilog ( -6+0.6 ) = [+H]

Antilog (-6) × Antilog (0.6) = [+H]

$Antilog (-6) = {10}^{ - 6} , \\ Antilog (0.6) = 4$

put the value in equation

${10}^{ - 6} \times 4 = [H+] \\ 4 \times {10}^{ - 6} = [H+]$

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2 years ago

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the standard change in Gibbs free energy is Δ????°′=7.53 kJ/molΔG°′=7.53 kJ/mol . Calculate Δ????ΔG for this reaction at 298 K29

Vera_Pavlovna [14]

Answer:

ΔG = 16.218 KJ/mol

Explanation:

dihydroxyacetone phosphate ↔ glyceraldehyde-3-phosphate

ΔG = ΔG° - RT Ln Q

∴ ΔG° = 7.53 KJ/mol * ( 1000 J / KJ ) = 7530 J/mol

∴ R = 8.314 J/K.mol

∴ T = 298 K

∴ Q = [glyceraldehyde-3-phosphate] / [dihydroxyacetone phosphate]

⇒ Q = 0.00300 / 0.100 = 0.03

⇒ ΔG = 7530J/mol - (( 8.314 J/K.mol) * ( 298 K ) * Ln ( 0.03 ))

⇒ ΔG = 16217.7496 J/mol ( 16.218 KJ/mol )

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4 years ago

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attashe74 [19]

Answer: sorry for the late answer, I just took the test today.

It provides instructions for processes of the cells

Explanation:

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4 years ago

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