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Tatiana [17]
3 years ago
11

Molecules of sugar are attracted to each other and form crystals. If you place sugar in water, the crystals break apart, but the

molecules remain the same kind of sugar. What can you infer about the bonds between atoms in a sugar molecule compared with the forces that attract sugar molecules to each other?
Chemistry
1 answer:
suter [353]3 years ago
3 0

Answer:

Explanation:

So here's the thing . Sugar molecule is a molecular solid . It a relatively weak molecule

We all know the formula of glucose ( which is a sugar type molecule ) which is C12 H22 O 11 . This is basically bonded together by dipole dipole forces ( remember that this force of attraction is weak ) .on the other hand you have the solution ( which is usually water) .water is held by strong attraction forces known as hydrogen bonding .since this type of bonding is greater than dipole dipole forces .therefore when glucose is dropped in water . water is easily able to obercome the forces that are keeping the glucose molecule intact . therefore the molecule dissolves and breaks into pieces. Hope this helps

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Potassium Chlorate decomposes according to the reaction below.
lesya692 [45]

Answer:

1.75L

Explanation:

Reaction of decomposition is:

2KClO₃(s) →  2KCl(s) + 3O₂(g)

We determine moles of salt:

4.35 g . 1 mol /122.55 g = 0.0355 moles

Ratio is 2:3. 2 moles of salt can produce 3 moles of oxygen

Then, our 0.0355 moles of chlorate may produce (0.0355 . 3)/ 2 = 0.0532 moles.

We have determined, moles of gas and we have data of pressure and temperature. To find out the volume, we apply the Ideal Gases Law:

We convert T° from °C to K → 27°C + 273 = 300K

P . V = n . R . T

0.75 atm . V = 0.0532 mol . 0.0821 L.atm/mol.K . 300K

V = (0.0532 mol  . 0.0821 L.atm/mol.K . 300K) / 0.75 atm

V  = 1.75 Liters

7 0
3 years ago
What type of change occurs when water changes from a solid to a liquid? A. a phase change B. a physical change C. an irreversibl
mars1129 [50]
B, a physical change because it is not destroying the substance.
3 0
3 years ago
Calculate the concentration of acetic acid, HAc, and acetate ion, Ac−, in a 0.25M acetate buffer solution with pH = 5.36. "0.25M
Neporo4naja [7]

Answer:

[HAc] = 0.05M

[Ac⁻] = 0.20M

Explanation:

The Henderson-Hasselbalch formula for the acetic acid buffer is:

pH = pka + log₁₀ [Ac⁻] / [HAc]

Replacing:

5.36 = 4.76 + log₁₀ [Ac⁻] / [HAc]

3.981 = [Ac⁻] / [HAc] <em>(1)</em>

Also, as total concentration of buffer is 0.25M it is possible to write:

0.25M =  [Ac⁻] + [HAc] <em>(2)</em>

Replacing (2) in (1)

3.981 = 0.25M - [HAc] / [HAc]

3.981 [HAc] = 0.25M - [HAc]

4.981 [HAc] = 0.25M

<em>[HAc] = 0.05M</em>

Replacing this value in (2):

0.25M =  [Ac⁻] + 0.05M

<em>[Ac⁻] = 0.20M</em>

I hope it helps!

7 0
4 years ago
How many seconds are in a 5.5 ms
deff fn [24]

Answer:

0.0055 seconds

Explanation:

To answer this question you need to know the equivalence:

1 second (s) = 5.5 (miliseconds) (ms)

Then you use the conversion factor:

-Put the value given and the units:

5.5ms

- Put in the denominator the equivalence with the same units:

5.5ms*\frac{}{1000ms}

- Put in the numerator the unit at which you want to convert the value given:

5.5ms*\frac{1s}{1000ms}

- Then multiply and divide the quantities:

5.5ms*\frac{1s}{1000ms}=0.0055 s

Note that the same units in numerator and denominator are cancelled.

8 0
3 years ago
What is the pH of an acetic acid solution where the concentration of acetic acid is 2 mM and the concentration of sodium acetate
schepotkina [342]

Answer:

pH = 5.76

Explanation:

We can solve this problem by using<em> Henderson-Hasselbach's equation</em>:

pH = pKa + log\frac{[SodiumAcetate]}{[AceticAcid]}

We are already know all the required information, thus we<u> input the data given by the problem</u>:

pH = 4.76 + log(20/2)

And finally <u>calculate the pH</u>:

pH = 5.76

The pH of that acetic acid solution is 5.76.

6 0
3 years ago
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