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hoa [83]
2 years ago
7

The diagram below shows the structures of a plant cell. What is structure 1, and how does it help sustain life? A. Structure 1 i

s a vacuole, which temporarily stores wastes. B. Structure 1 is the cell wall, which provides structural support for the cell. C. Structure 1 is the nucleus, which is the control center of the cell. D. Structure 1 is a chloroplast, which is the site of photosynthesis.
Chemistry
2 answers:
Monica [59]2 years ago
6 0

Answer:

D Structure 1 is a chloroplast which is the site of photosynthesis.

Explanation:

I got it correct on my test

Aneli [31]2 years ago
4 0

From the diagram, structure 1 would be the chloroplast and it represents the site of photosynthesis.

<h3>Plant cells</h3>

The image (see attached) is that of a plant cell and structure labeled 1 is the chloroplast.

The chloroplast of plant cells acts as the site of photosynthesis where inorganic materials are used to synthesize carbohydrates.

Plants generally serve as producers in food chains and thus, chloroplast helps sustain life.

More on plant cells can be found here: brainly.com/question/1894299

You might be interested in
Please help me and explanation would be really awesome thank you!
valentina_108 [34]

Answer:

So the answer would be 10 moles

Explanation:

1) Start with the molecular formula for water: H_{2} O!

2) If there are 10 moles of water use a mole ratio to calculate the moles of oxygen it would produce.

(This question is... interesting... since they chose an element that is diatomic in free state so It could TECHNICALLY be two answers, moles of O or moles of O_{2})

The mole ratio is 1 moles of H_{2}O to 1 moles of O. This is because the coefficient for oxygen in water is simple 1, so the ratio is 1:1.

3) that means if 10 moles of water decompose, they decompose into 10 moles of H_{2} and 10 moles of O.

Extra:

About what I was saying before about the question being slightly interesting:

10 moles of pure oxygen is produced but free state oxygen exists as O_{2} so it could possibly be 10 OR 5! However, notice it says elements. This leads me to believe the answer is 10 (monatomic oxygen) instead of 5 (free state/diatomic oxygen).

I hope this helps!

4 0
2 years ago
½H2(g) + ½I2(g) → HI(g), ΔH = +6.2 kcal/mole 21.0 kcal/mole + C(s) + 2S(s) → CS2(l) What type of reaction is represented by the
wlad13 [49]

<u>Answer:</u>

Exothermic Reaction are those reaction, in which energy is released while in endothermic reaction are those, in which energy is absorbed.

<u>Explanation:</u>

First Reaction:

As in this reaction, energy is released

½H2(g) + ½I2(g) → HI(g), ΔH = +6.2 kcal/mole

so it is <em>exothermic reaction</em>

Second reaction:

As in this reaction, energy is absorbed

21.0 kcal/mole + C(s) + 2S(s) → CS2(l)

so it is <em>endothermic reactions</em>.


7 0
3 years ago
Read 2 more answers
In principle, the equilibrium in the dehydration of an alcohol could be shifted to the right be removal of water. Why is this ta
Trava [24]

Answer:

See explanation

Explanation:

In this case, we have to remember that if we want to remove water from the reaction vessel we have to heat the vessel. So, we can convert the liquid water into <u>gas water</u> and we can remove it from the vessel. In this case, the products of dehydration for both molecules are <u>(E)-4-methylpent-2-ene</u> and <u>cyclohexene</u> with boiling points of <u>59.2 ºC</u> and <u>89 ºC</u> respectively. The boiling point of water is <u>100 ºC</u>, therefore if we heat the vessel the products and water would leave the system, and the products would be lost.

See figure 1

I hope it helps!

3 0
3 years ago
Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho
Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3

c) The leftover is computed as follows:

m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0
2 years ago
Why are valence electrons important?
Usimov [2.4K]
They are the outer layer of the electron layers.
5 0
3 years ago
Read 2 more answers
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