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Studentka2010 [4]
3 years ago
5

What two structures would provide a positive identification of a plant cell under a microscope? a)cell wall, mitochondria b)plas

ma membrane, large central vacuole c)ribosomes, chloroplasts d)cell wall, chloroplast
Chemistry
2 answers:
lutik1710 [3]3 years ago
6 0

<h2>Answer:</h2>

The correct answer is option D which is cell wall, chloroplast.

<h3>Explanation:</h3>
  • Here positive identification means such features in plant cell which are significantly distinguishes plant cell from animals cells.
  • Cell wall is outer envelope type structure present around the cell membrane. It is not present in animal cells.
  • Chloroplasts are plasmids. These organelles are distinguished structure of photosynthesizing cells.
  • Hence only plant cells have chloroplast while chloroplast are absent i animal cells.
  • So from looking at chloroplast and cell wall we can identify a plant cell.
olga55 [171]3 years ago
3 0

Cell wall, Large central vacuole, Chloroplasts

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The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way:
LiRa [457]

Explanation:

(a)  As the given chemical reaction equation is as follows.

           OCl^{-} + I^{-} \rightarrow OI^{-1} + Cl^{-1}

So, when we double the amount of hypochlorite or iodine then the rate of the reaction will also get double. And, this reaction is "first order" with respect to hypochlorite and iodine.

Hence, equation for rate law of reaction will be as follows.

              Rate = K \times [OCl^{-}] \times [l^{-}]

(b)  Since, the rate equation is as follows.

                    Rate = K [OCl^{-}][l^{-}]

Let us assume that ([OCl^{-}] = [l^{-}])

Putting the given values into the above equation as follows.

             1.36 \times 10^{-4} = K \times (1.5 \times 10^{-3})^2

            1.36 \times 10^{-4} = K \times (2.25 \times 10^{-6})

                   K = \frac{1.36 \times 10^{-4}}{2.25 \times 10^{-6}}

                      = 60.4 M^{-1}sec^{-1}

Hence, the value of rate constant for the given reaction is 60.4 M^{-1}sec^{-1} .

(c) Now, we will calculate the rate as follows.

                Rate = K [OCl^{-}][l^{-}]

                         = 60.4 \times (1.8 \times 10^{3}) \times (6.0 \times 10^{4})

                        = 6.52 \times 10^{5}

Therefore, rate when [OCl^{-}] = 1.8 \times 10^{3} M and [I^{-}]= 6.0 \times 10^{4} M is  6.52 \times 10^{5}.

8 0
2 years ago
After he conducted cathode ray tube experiments proving the existence of negatively charged particles we now call electrons, Tho
Lina20 [59]

Answer:

Answer is explained below;

Explanation:

In 1904, after the discovery of the electron, the English physicist Sir J.J. Thomson proposed the plum pudding model of an atom. In this model, the atom had a positively-charged space with negatively charged electrons embedded inside it i.e., like a pudding (positively charged space) with plums (electrons) inside.

In 1911, another physicist Ernest Rutherford proposed another model known as the Rutherford model or planetary model of the atom that describes the structure of atoms. In this model, the small and dense atom has a positively charged core called the nucleus. Also, he proposed that just like the planets revolving around the Sun, the negatively charged electrons are moving around the nucleus.

By conducting a gold foil experiment, Rutherford disproved Thomson's model. In this experiment, positively charged alpha particles emitted from a radioactive source enclosed within a protective lead were used which was then focused into a narrow beam. It was then passed through a slit in front of which a thin section of gold foil was placed. A fluorescent screen (coated with zinc sulfide) was also placed in front of the slit to detect alpha particles which on striking the fluorescent screen would produce scintillation (a burst of light) which was visible through a microscope attached to the back of the screen.

He observed that most of the alpha particles passed straight through the gold foil without any resistance and this implied that atoms contain a large amount of open space. The slight deflection of some of the alpha particles, the large-angle scattering of other alpha particles and even the bouncing back of a very few alpha particles toward the source suggested their interactions with other positively charged particles inside the atom.

So, he concluded that only a dense and positively charged particle such as the nucleus would be responsible for such strong repulsion. Also, the negatively charged electrons electrically balanced the positive nuclear charge and they moved around the nucleus in circular orbits. Between the electrons and nucleus, there was an electrostatic force of attraction just like the gravitational force of attraction between the sun and the revolving planets.

Later, the Rutherford model was replaced by the Bohr atomic model.

6 0
3 years ago
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lukranit [14]

Answer:

C. water acts as a radiation shield to reduce the radiation level

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Pavel [41]

Answer:

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Explanation:

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The "particles" of a gas are actually protons or electrons. true or false
Andrew [12]
False.

Hope this helps, Good luck on the assignment.
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