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3 years ago

why do you think Kool Aid comes in packets that has a very finely ground powder (finely ground means very small particles).

Chemistry

1 answer:

marishachu [46]3 years ago

8 0

Answer:

Answer 1:

When you pour the kool-aid into water, the little crystals go straight to the bottom because they are heavier than the water. If you left them there without stirring, and came back a few days later, you wouldn't see any crystals on the bottom. That's because the stuff in kool-aid can DISSOLVE in water, which means that each little molecule of kool-aid gets suspended between the molecules of water. When that happens, you can't see the kool-aid anymore...it's trapped between the water molecules. When you stir kool-aid, you help DISSOLVE the kool-aid in water by keeping all of the crystals off the bottom and in the water. So you see, stirring kool-aid speeds up the dissolving,

Answer 2:

Are you referring to Koolaid in the granular form?If so the koolaid grains sink in water because the grains have a greater density than that of water. Once your stir the grains dissolve and go into solution where they remain because the dissolved koolaid is miscible with water unlike oil (floats) or gasoline (sinks). How long did you let the koolaid remain in the water before you stirred it? I would think that if you left it undisturbed for a long time (days) it would eventually mix on its own.

Answer 3:

I'm not a chemist, but I think I can answer your question about Kool-Aid. Kool-Aid is mostly sugar, which is heavier than water, so when you pour it in it sinks to the bottom. When you stir it up the sugar (and flavoring) dissolves so that you don't have any solid particles any more. Stuff that is dissolved in water will not sink because it is no longer a physically separate thing. It becomes part of the water (or water-sugar-flavor solution). What happens if you pour the Kool-Aid in but don't stir it? Will it eventually dissolve? You may have to wait a long time, like over night. Try it and let me know what you find!

Answer 4:

It all has to do with the rate at which kool-aid crystals (basically its SUGAR!!) dissolves in water relative to the rate at which the sugar crystals sink. If you just dump the stuff in, it sinks because it is denser than the water. As it sinks it dissolves. But when you stir the water, the rate of dissolution becomes greater than the rate of sinking and so the crystals dissolve before they reach the bottom. So it all has to do with the comparison between the rate of sinking versus the rate of dissolution.

Now I have an experiment for you. What happens if you mix up some Jello and instead of letting it sit still, you keep stirring it??? WILL THE JELLO EVER SET??

You may have to borrow your mom's mixing machine because you will get tired of stirring after 10 minutes!!!!

If you do the experiment let me know how it turns out. Actually, you should set up a control. Make two batches of Jello...with one, put it in the refrigerator and dont stir; with the other, keep stirring it (in the refrigerator), if you can figure how to arrange that without your mom or dad getting mad!!!

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A certain substance melts at a temperature of . But if a sample of is prepared with of urea dissolved in it, the sample is found

pshichka [43]

Answer:

2.2 °C/m

Explanation:

It seems the question is incomplete. However, this problem has been found in a web search, with values as follow:

" A certain substance X melts at a temperature of -9.9 °C. But if a 350 g sample of X is prepared with 31.8 g of urea (CH₄N₂O) dissolved in it, the sample is found to have a melting point of -13.2°C instead. Calculate the molal freezing point depression constant of X. Round your answer to 2 significant digits. "

So we use the formula for <em>freezing point depression</em>:

ΔTf = Kf * m

In this case, ΔTf = 13.2 - 9.9 = 3.3°C

m is the molality (moles solute/kg solvent)

350 g X ⇒ 350/1000 = 0.35 kg X

31.8 g Urea ÷ 60 g/mol = 0.53 mol Urea

Molality = 0.53 / 0.35 = 1.51 m

So now we have all the required data to <u>solve for Kf</u>:

ΔTf = Kf * m

3.3 °C = Kf * 1.51 m

Kf = 2.2 °C/m

5 0

2 years ago

A compound contains only change and n combustion of 35.0mg of the compound produces 33.5mg co2 and 41.1mg h2o. What is the empir

Viefleur [7K]

Answer:

The empirical formula is CH6N2

Explanation:

A compound containing only C, H, and N yields the following data. Complete combustion of 35.0 mg of the compound produced 33.5 mg of CO2 and 41.1 mg of H2O. What is the empirical formula of the compound

Step 1: Data given

Mass of the compound = 35.0 mg = 0.035 grams

Mass of CO2 = 33.5 mg = 0.0335 grams

Mass of H2O = 41.1 mg = 0.0411 grams

Molar mass CO2 = 44.01 g/mol

Molar mass H2O = 18.02 g/mol

Molar mass C = 12.01 g/mol

Molar mass O = 16.0 g/mol

Molar mass H = 1.01 g/mol

Step 2: Calculate moles CO2

Moles CO2 = 0.0335 grams / 44.01 g/mol

Moles CO2 = 7.61 *10^-4 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 7.61 *10^-4 moles CO2 we have 7.61 *10^-4 moles C

Step 4: Calculate mass C

Mass C = 7.61 *10^-4 moles * 12.01 g/mol

Mass C = 0.00914 grams = 9.14 mg

Step 5: Calculate moles H2O

Moles H2O = 0.0411 grams / 18.02 g/mol

Moles H2O = 0.00228 moles

Step 6: Calculate moles H

For 1 mol H2O we have 2 moles H

For 0.00228 moles H2O we have 2* 0.00228 = 0.00456 moles H

Step 7: Calculate mass H

Mass H = 0.00456 moles * 1.01 g/mol

Mass H = 0.00461 grams = 4.61 mg

Step 8: Calculate mass N

Mass N = 35.0 mg - 9.14 - 4.61 = 21.25 mg = 0.02125 grams

Step 9: Calculate moles N

Moles N = 0.02125 grams / 14.0 g/mol

Moles N = 0.00152 moles

Step 10: Calculate the mol ratio

We divide by the smallest amount of moes

C: 0.000761 moles / 0.000761 moles= 1

H: 0.00456 moles / 0.000761 moles = 6

N: 0.00152 moles / 0.000761 moles = 2

For every C atom we have 6 H atoms and 2 N atoms

The empirical formula is CH6N2

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https://smartlyhelper.com/chemistry/question14062929

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2 years ago

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3 years ago

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Heat is transferred by conduction through a wall of thickness 0.87 m. What is the rate of heat transfer if the walls thermal con

erik [133]

Explanation:

The given data is as follows.

Thickness (dx) = 0.87 m, thermal conductivity (k) = 13 W/m-K

Surface area (A) = 5 $m^{2}$ , $T_{1} = 14^{o}C$

$T_{2} = 93^{o}C$

According to Fourier's law,

Q = $-kA \frac{dT}{dx}$

Hence, putting the given values into the above formula as follows.

Q = $-kA \frac{dT}{dx}$

= $-13 W/m-k \times 5 m^{2} \times \frac{(14 - 93)}{0.87 m}$

= 5902.298 W

Therefore, we can conclude that the rate of heat transfer is 5902.298 W.

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3 years ago