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2 years ago

Sugar cannot be separated from sugar solution by filtering explain why?

Chemistry

2 answers:

luda_lava [24]2 years ago

7 0

Answer:

Sugar can not be separated due it has very fine particles.

Explanation:

When we use to separate sugar with salt they both are dissolved with the water.

alexdok [17]2 years ago

5 0

Answer:

The sugar particles are so small that they have dissolved in the water, and can easily pass through the filter.

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Help! ASAP! Please on the questions

KiRa [710]

Answer:

62.15 m (squared)

Explanation:

We know that the Area of the square is 5.5 x 4= 22

so 90 degree to 30 degree is divide by 3

so u divide 22 by 3 and get 7.3 wich is the semi circle thing

now 7.3 x 5.5 = 40.15

22 + 40.15 = 62.15 m (squared)

Look im not really sure but I think this is the answer

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8 0

2 years ago

Read 2 more answers

Why are there 2 chloride ions for every 1 calcium in calcium chloride

tensa zangetsu [6.8K]

Answer:

See explanation

Explanation:

Calcium is divalent. This means that it donates two electrons during ionic bond formation. Since chlorine atom can only accept one electron during ionic bond formation, two chlorine atoms must accept the two electrons donated by calcium.

For this purpose, each time CaCl2 is formed, there must be two chlorine atoms for each calcium atom.

5 0

2 years ago

While ethanol (CH3CH2OH) is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by

grin007 [14]

Answer:

$n_{C2H_5OH}^{eq}=14.234mol$

Explanation:

Hello,

In this case, the reaction is:

$C_2H_4+H_2O\rightleftharpoons CH_3CH_2OH$

Thus, the law of mass action turns out:

$Kc=\frac{[CH_3CH_2OH]_{eq}}{[H_2O]_{eq}[CH_2CH_2]_{eq}}$

Thus, since at the beginning there are 29 moles of ethylene and once the equilibrium is reached, there are 16 moles of ethylene, the change $x$ result:

$[CH_2CH_2]_{eq}=29mol-x=16mol\\x=29-16=13mol$

In such a way, the equilibrium constant is then:

$Kc=\frac{\frac{x}{V} }{\frac{16mol}{V}* \frac{3mol}{V}} =\frac{\frac{13mol}{75.0L} }{\frac{16mol}{75.0L}* \frac{3mol}{75.0L}} =20.31$

Thereby, the initial moles for the second equilibrium are modified as shown on the denominator in the modified law of mass action by considering the added 15 moles of ethylene:

$Kc=\frac{\frac{13+x_2}{V} }{\frac{16+15-x_2}{V}* \frac{3-x_2}{V}} =20.31$