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krok68 [10]
3 years ago
13

PLEASE SOMEONE HELP ASAP: why does alcohol dissolve permanent marker

Chemistry
1 answer:
Volgvan3 years ago
4 0

Explanation:

Isopropyl alcohol is the solvent used in whiteboard markers. When it comes into contact with the permanent marker, the solvent breaks the cohesive force between the ink molecules and the whiteboard by solubilising them (i.e. returning them to solution).

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A single 250 gram
Gala2k [10]

2.2 x 10^-2

0.055 / 250 = 0.00022 - This would be 2.2 x 10^-4, but the question is asking for percent, not proportion, so multiply by 100% to get the percentage.

0.00022 * 100% = 0.022% = 2.2 * 10^-2

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3 years ago
As the tempreture of a liquid increases the solubility of a liquid of that liquid
anzhelika [568]
As the temperature increases, the solubility of the solute in the liquid also increases.  This is due to the fact that the increase in energy allows the liquid to more effectively break up the solute.  The additoin of energy also shifts the equilibrium of the reation to the right since it takes energy to dissolve most things and you are adding more of it (this is explained with Le Chatlier principles).

I hope this helps and also I assumed that your question involved the solubility of an ionic substance in a solvent like water.  If that was not your question feel free to say so in the comments so that I can answer your actually question.
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3 years ago
What is HC2H3O2 in science
Leona [35]
The chemical name of Hc2h3o2 is Acetic Acid.
3 0
3 years ago
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A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s
Alla [95]

<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For NaBr:</u>

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol

The chemical equation for the reaction of 1-butanol and NaBr is:

\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = \frac{1}{1}\times 0.108=0.108 moles of 1-bromobutane

  • Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g

  • To calculate the percentage yield of 1-bromobutane, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%

Hence, the percent yield of the 1-bromobutane is 48.65 %

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3 years ago
1. The side-by-side overlap of p orbitals produces what kind of bond?
Dmitry_Shevchenko [17]
A pi bond... hope this helps!!!!!
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3 years ago
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