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andrey2020 [161]
2 years ago
14

Use the drop-down menus to name these structures F FICH H

Chemistry
1 answer:
miss Akunina [59]2 years ago
3 0

Answer:

What Structures??

Explanation:

Can you please explain the question more

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The right answer for the question that is being asked and shown above is that: "Water’s polarity produces a high density, which allows water to move to the leaves." Example best shows that the chemistry of water is helpful to plants is that <span>Water’s polarity produces a high density, which allows water to move to the leaves.</span>
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3 years ago
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¿Cuál era su juego favorito cuando eran niños?​
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Explanation:

7 0
3 years ago
Which of the following statements is true about the specific heat capacity of a substance?
Ludmilka [50]

Answer:

Option A

Explanation:

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Physical properties can be observed or measured without changing the composition of matter. Physical properties are used to observe and describe matter. Physical properties include: appearance, texture, color, odor, melting point, boiling point, density, solubility, polarity, specific heat  and many others.

6 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
3 years ago
Any GUY wanna talk im bore.d
harina [27]

\boxed {\boxed {Answer}}

ⁿᵘ ᵗʰˣ ⁱᵐ ᵃ ᵍᵘʳˡ -.-

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