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Sloan [31]
3 years ago
6

5) O etanol C2H6O, combustível produzido em grande escala no Brasil, pode ser obtido pela fermentação da sacarose C12H22O11 enco

ntrada na cana-de-açúcar. Esse processo consiste na degradação por enzimas de micro-organismos da sacarose. Um técnico de laboratório preparou uma solução aquosa misturando 33 g de sacarose em 200 mL de água. Ao final, a solução ficou com um volume de 220 mL. Qual a concentração em g/l desta solução sabendo que a formula é C=Mg /VL a) 0,15 b) 0,165 c) 66 d) 15 e) 150
Chemistry
1 answer:
UNO [17]3 years ago
3 0

Answer:

a) 0.15

Explanation:

A concentração de sacarose é 0,15 g / l no etanol. O Brasil está produzindo etanol em grandes quantidades que é usado como combustível em veículos e indústrias. Este etanol é produzido a partir da cultura da cana-de-açúcar e o Brasil é o segundo maior produtor mundial com cerca de 34,45 bilhões de litros em 2019. Os Estados Unidos da América e o Brasil produzem cerca de 84% do etanol mundial.

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Answer:

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Explanation:

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How many moles of water react with 0.946 mol of nitrogen dioxide
VARVARA [1.3K]
<span>the answer is
2.7 moles

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How many spatial orientations are available for the p sublevel
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Do your cells have access to everything they need over the first 4 or 5 hours that you are lost? Yes or No?
77julia77 [94]

Explanation: cell in the body is enclosed by a cell (Plasma) membrane. The cell membrane separates the material outside the cell, extracellular, from the material inside the cell, intracellular. ... All materials within a cell must have access to the cell membrane (the cell's boundary) for the needed exchange

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4 0
3 years ago
If you combine 230.0 mL 230.0 mL of water at 25.00 ∘ C 25.00 ∘C and 120.0 mL 120.0 mL of water at 95.00 ∘ C, 95.00 ∘C, what is t
Thepotemich [5.8K]

<u>Answer:</u> The final temperature of the mixture is  49°C

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

  • <u>For cold water:</u>

Density of cold water = 1 g/mL

Volume of cold water = 230.0 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{230.0mL}\\\\\text{Mass of water}=(1g/mL\times 230.0mL)=230g

  • <u>For hot water:</u>

Density of hot water = 1 g/mL

Volume of hot water = 120.0 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{120.0mL}\\\\\text{Mass of water}=(1g/mL\times 120.0mL)=120g

When hot water is mixed with cold water, the amount of heat released by hot water will be equal to the amount of heat absorbed by cold water.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]      ......(1)

where,

q = heat absorbed or released

m_1 = mass of hot water = 120 g

m_2 = mass of cold water = 230 g

T_{final} = final temperature = ?°C

T_1 = initial temperature of hot water = 95°C

T_2 = initial temperature of cold water = 25°C

c = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

120\times 4.186\times (T_{final}-95)=-[230\times 4.186\times (T_{final}-25)]

T_{final}=49^oC

Hence, the final temperature of the mixture is  49°C

4 0
3 years ago
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