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Maksim231197 [3]

3 years ago

7

During mitosis, this structure moves individual chromosomes?

1 answer:

stiv31 [10]3 years ago

3 0

Answer:

Anaphase

Explanation:

Two separate classes of movements occur during anaphase. During the first part of anaphase, the kinetochore microtubules shorten, and the chromosomes move toward the spindle poles. During the second part of anaphase, the spindle poles separate as the non-kinetochore microtubules move past each other.

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This is my question for science 9:

kvasek [131]

Yes, she is right. 1: when she took them out of the freezer heat was added. 2: when she toasted them in the oven heat was added and a chemical change occurred because the baking soda/powder reacted, causing the muffins to rise.

5 0

3 years ago

Which statement provides background knowledge that helps a reader understand the theme of valuing all life found in"Birdfoot's G

alekssr [168]

Answer:

i think its A

Explanation:

7 0

4 years ago

Read 2 more answers

Calculate the standard cell potential (E∘) for the reaction X(s)+Y+(aq)→X+(aq)+Y(s) if K = 8.97×10−3. Express your answer to thr

puteri [66]

Answer: The standard cell potential (E∘) for the reaction $X(s)+Y^+(aq)\rightarrow X^+(aq)+Y(s)$ is -0.121 V

Explanation:

The reaction is:

$X(s)+Y^+(aq)\rightarrow X^+(aq)+Y(s)$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K$

where,

$\Delta G^o$ = Standard Gibbs free energy = ?

R = Gas constant = $8.314J/K mol$

T = temperature = 298 K

K= equilibrium constant = $8.97\times 10^{-3}$

Putting values in above equation, we get:

$\Delta G^0=-(8.314J/Kmol\times 298K\times \ln (8.97\times 10^{-3})\\\\\Delta G^0=11678.9J/mol$

To calculate standard Gibbs free energy, we use the equation:

$\Delta G^o=-nFE^o_{cell}$

Where,

n = number of electrons transferred = 1

F = Faradays constant = 96500 C

$E^o_{cell}$ = standard cell potential = ?

Putting values in above equation, we get:

$11678.9J/mol=-1\times 96500\times E^0_{cell}$

$\frac{11678.9J/mol}{-96500}=E^0_{cell}$

$-0.121V=E^0_{cell}$

Thus standard cell potential (E∘) for the reaction $X(s)+Y^+(aq)\rightarrow X^+(aq)+Y(s)$ is -0.121 V

5 0

4 years ago

The decomposition of dinitrogen pentoxide, N2O5, to NO2 and O2 is a first-order reaction. At 60°C, the rate constant is 2.8 × 10

Ugo [173]

Answer:

The correct option is a.

Explanation:

$2N_2O_5\rightarrow 4NO_2 + O_2$

125 kPa

125kpa - 2x 4x x

Total pressure after reaction = 176 kPa

125 kPa - 2x + 4x + x = 176 kPa

x = 17

125 kpa - 2x = 125 kPa - 2(17) = 91 kPa

Initial pressure of the dinitrogen pentoxide ,(at t=0) = $P_o= 125 kPa$

Final pressure of the dinitrogen pentoxide, (at t = t) = P = 91 kPa

The rate constant is = $k = 2.8\times 10^{-3} min^{-1}$

$t=\frac{2.303}{k}\log\frac{P_o}{P}$

$t=\frac{2.303}{2.8\times 10^{-3} min^{-1}}\log\frac{125 kPa}{91 kPa}$

$t=113.3969 minutes\approx 113 minutes$

It will take 113 minutes for the total pressure to reach 176 kPa.

8 0

3 years ago

Which statement best describes an example of selective breeding?

Gnoma [55]

Answer:

B people control how plants or animals reproduce to get desired traits

Explanation:

if you get it correct mark meh brainliest or however you say it so ppl know its correct

3 0

3 years ago