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3 years ago

Identify the reactant that gets oxidized in the following reaction.

Chemistry

1 answer:

Nat2105 [25]3 years ago

6 0

Answer:

E) This is not an Oxidation/ Reduction type of reaction

Explanation:

Oxidation involves the loss of electrons while reduction involves the addition of electrons. None of the elements change their oxidation number in the reaction, so this is not an oxidation/reduction type of reaction.

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

tankabanditka [31]

Answer: D. CYTOKINESIS

Explanation:

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

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3 years ago

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Lena [83]

Answer:

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3 years ago

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mrs_skeptik [129]

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3 years ago

Which gas is least likely to obey the ideal gas laws at very high pressures and very low temperatures?

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4 years ago

Calculate the maximum KE and velocity of an electron from zinc by a 275nm photon

Dmitry [639]

Answer:

Kinetic energy: $6.024*10^{-20}$

Velocity: $3.64*10^{5}$

Explanation:

From the equations of the photo-electric effect,

We know:

$hv=hv_0+K.E$

Where,

1. $h$ is the Planck's constant which is $6.626*10^{-34}$

2. $v,v_0$ are the frequency of light emitted and threshold frequencies respectively.

3. $K.E$ is the kinetic energy of the electrons emitted.

By fact, we come to know that the threshold frequency of Zn is 300nm

And also $v=\frac{c}{d}$

Where ,

1. $c$ is the speed of light $=3*10^8$

2. $d$ is the wavelength.

Thus,

$\frac{hc}{d}=\frac{hc}{d_0}+K.E\\K.E=\frac{hc}{d}-\frac{hc}{d_0}\\K.E=hc*(\frac{1}{d}-\frac{1}{d_0})\\K.E=6.626*10^{-34}*3*10^8*(\frac{1}{275*10^{-9}}-\frac{1}{300*10^{-9}})\\K.E=6.024*10^{-20}kgm^2s^{-2}$

Now to find velocity:

$K.E=\frac{1}{2}mv^2\\v^2=1.324*10^{11}\\v=3.64*10^5$

8 0

3 years ago