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Sliva [168]
4 years ago
13

What structure holds the sister chromatids to the spindle fibers?

Chemistry
1 answer:
olganol [36]4 years ago
8 0
The answer is <span>a. kinetochore.

A kinetochore is a protein structure that holds the </span><span>sister chromatids to the spindle fibers. It is the place on chromatids where the spindle fibers bind during the cell division. As the result, sister chromatids are pulled apart to the opposite ends of the cell.</span>
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Liquid<br><br> Describe this state of matter in 3-4 sentences
Law Incorporation [45]

Answer: Liquid has no definite shape but it has definite volume. The particles are free to move over each other but are still attracted to each other. Liquids can be compressed into gas.

Hope this helps :)

4 0
3 years ago
PLEASE HELP 20 points
777dan777 [17]

  n = 1.5atm (15L) / .0821 (280k) = .98 mol NaCl

  NaCl = 22.99g Na + 35.45g Cl = 58.44g NaCl

  58.44g NaCl x .98 mol NaCl = 57.27g NaCl

Explanation:

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4 0
3 years ago
What is the answer?
Dmitry_Shevchenko [17]

as a heterogeneous mixture

Explanation:

because I just know that it is

6 0
4 years ago
Solid NaBr is slowly added to a solution that is 0.073 M in Cu+ and 0.073 M in Ag+.Which compound will begin to precipitate firs
saul85 [17]

Answer :

AgBr should precipitate first.

The concentration of Ag^+ when CuBr just begins to precipitate is, 1.34\times 10^{-6}M

Percent of Ag^+ remains is, 0.0018 %

Explanation :

K_{sp} for CuBr is 4.2\times 10^{-8}

K_{sp} for AgBr is 7.7\times 10^{-13}

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgBr has a smaller than CuBr then AgBr should precipitate first.

Now we have to calculate the concentration of bromide ion.

The solubility equilibrium reaction will be:

CuBr\rightleftharpoons Cu^++Br^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Cu^+][Br^-]

4.2\times 10^{-8}=0.073\times [Br^-]

[Br^-]=5.75\times 10^{-7}M

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

AgBr\rightleftharpoons Ag^++Br^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Ag^+][Br^-]

7.7\times 10^{-13}=[Ag^+]\times 5.75\times 10^{-7}M

[Ag^+]=1.34\times 10^{-6}M

Now we have to calculate the percent of Ag^+ remains in solution at this point.

Percent of Ag^+ remains = \frac{1.34\times 10^{-6}}{0.073}\times 100

Percent of Ag^+ remains = 0.0018 %

3 0
3 years ago
Which of the following energy sources are in some way divided from the sun (select all that apply)
alukav5142 [94]
The correct answer is b
3 0
4 years ago
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