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IrinaVladis [17]
3 years ago
11

What Causes Surface Tension?

Chemistry
1 answer:
Nostrana [21]3 years ago
4 0

Answer:

The cohesive nature of the molecules

Explanation:

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CAN ANYONE ANSWER THIS ONE CHEMISTRY QUESTION PLZZZ!!!
xz_007 [3.2K]

Answer:

2.8 oxygen

Explanation:

8 0
3 years ago
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What percentage of radium 288 is left after 1600 years ? ( figure out the number of half-lives,then look at the chart above to f
erastovalidia [21]

Answer:

Find it ur self lol

Explanation:

lol just look it up

5 0
3 years ago
Name the type of bond in organic chemistry that corresponds to a glycoside bond
Semenov [28]

Answer:

covalent bond

Explanation:

The bond which is most common in the organic molecules is the covalent bond which involves sharing of the electrons between the two atoms.

Glycosidic bond, also known as glycosidic linkage is type of the covalent bond which joins carbohydrate molecule to other group that may not or may be a carbohydrate.

Glycosidic bond is the bond which is formed between hemiacetal or hemiketal group of the saccharide and hydroxyl group of compounds like alcohol.

5 0
3 years ago
6.) (5 points) Assume you have a system with a fixed pH of 4.0. It is well buffered and therefore the pH will not change. What i
KIM [24]

Answer:

Dissociated state is the predominant one

Explanation:

When a molecule with pKa of 4.52 is in an aqueous solution at pH = 4.0, follows the H-H equation, thus:

pH = pKa + log₁₀ [A⁻] / [HA]

<em>Where [A⁻] is the dissociated state and [HA] represents the protonated state</em>

Replacing:

4.0 = 5.2 + log₁₀ [A⁻] / [HA]

-1.2 = log₁₀ [A⁻] / [HA]

0.063 =  [A⁻] / [HA]

[HA]  = 16 [A⁻]

That means you have 16 times more [HA] than [A⁻] and the <em>dissociated state is the predominant one</em>

7 0
3 years ago
What is the mass percent of sucrose (C12H22O11, Mm = 342 g/mol) in a 0.329-m sucrose solution?
Hitman42 [59]

Answer:

\% m/m=10.1\%

Explanation:

Hello,

In this case given the molal solution of sucrose, we can assume there are 0.329 moles of sucrose in 1 kg of solvent, thus, computing both the mass of sucrose and solvent in grams, we obtain:

m_{sucrose}=0.329mol*\frac{342g}{1mol}=112.5g

m_{solvent}=1000g

In such a way, we proceed to the calculation of the mass percent as follows:

\% m/m=\frac{112.5g}{112.5g+1000g}*100\%\\ \\\% m/m=10.1\%

Regards.

8 0
3 years ago
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