All categories

2 years ago

A stain is made up of molecules that do not have charged regions. What soap/solvent combination should I use?Are there multiple?

Chemistry

1 answer:

Dmitry_Shevchenko [17]2 years ago

4 0

Answer:

The answer is below

Explanation:

Yes, there are multiple combinations.

Firstly, a solvent or soap combination could be used as a polar solvent with soap. In this way, the nonpolar region of the soap molecules would mix directly with the stain while the solvent would surround the soap-stain micelle.

Secondly, another method or way is to have a combination of a nonpolar solvent with no soap. In this method, however, it is the solvent that would mix directly with the stain.

You might be interested in

combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon

masya89 [10]

Answer:

$\rm CH_2$ .

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be $\rm CO_2$ and $\rm H_2O$ . The $\rm C$ and $\rm H$ would come from the hydrocarbon, while the $\rm O$ atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

$\rm C$ : $12.011$ .
$\rm H$ : $1.008$ .
$\rm O$ : $\rm 15.999$ .

Calculate the molar mass of $\rm CO_2$ and $\rm H_2O$ :

$M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}$ .

$M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}$

Calculate the number of moles of $\rm CO_2$ molecules in $33.01\; \rm g$ of $\rm CO_2\!$ :

$\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol$ .

Similarly, calculate the number of moles of $\rm H_2O$ molecules in $13.51\; \rm g$ of $\rm H_2O\!$ :

$\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol$ .

Note that there is one carbon atom in every $\rm CO_2$ molecule. Approximately $0.7501\; \rm mol$ of $\rm CO_2\!$ molecules would correspond to the same number of $\rm C$ atoms. That is: $n(\mathrm{C}) \approx 0.7501\; \rm mol$ .

On the other hand, there are two hydrogen atoms in every $\rm H_2O$ molecule. approximately $0.7499\; \rm mol$ of $\rm H_2O$ molecules would correspond to twice as many $\rm H\!$ atoms. That is: $n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol$ .

The ratio between the two is: $n(\mathrm{C}): n(\mathrm{H}) \approx 1:2$ .

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be $\rm CH_2$ .

6 0

3 years ago

Small quantities of hydrogen can be prepared by the addition of hydrochloric acid to zinc. a sample of 195 ml of hydrogen was co

Vikentia [17]

Let's go over the given information. We have the volume, temperature and pressure. From the ideal gas equation, that's 4 out of 5 knowns. So, we actually don't need Pvap of water anymore. Assuming ideal gas, the solution is as follows:

PV=nRT
Solving for n,
n = PV/RT = (753 torr)(1 atm/760 torr)(195 mL)(1 L/1000 mL)/(0.0821 L·atm/mol·K)(25+273 K)
n = 7.897×10⁻³ mol H₂

The molar mass of H₂ is 2 g/mol.

Mass of H₂ = 7.897×10⁻³ mol * 2 g/mol = <em>0.016 g H₂</em>

5 0

3 years ago

Read 2 more answers

I don't really understand this worksheet question.

Citrus2011 [14]

25/2 and 96/X
CROSS MULTIPLY.

2x=2,400.
divide by 2.
x=1,200.

you take the GIVEN MASS of an element, and you put it on top, the coefficient is what it’s over. i believe this is right

3 0

3 years ago

The rate of effusion of an unknown gas was measured and found to be 11.9 mL/min. Under identical conditions, the rate of effusio

iren2701 [21]

Answer : The correct option is, (B) $CO_2$

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

$R\propto \sqrt{\frac{1}{M}}$

or,

$(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}$ ..........(1)

where,

$R_1$ = rate of effusion of unknown gas = $11.9\text{ mL }min^{-1}$

$R_2$ = rate of effusion of oxygen gas = $14.0\text{ mL }min^{-1}$

$M_1$ = molar mass of unknown gas = ?

$M_2$ = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the above formula 1, we get:

$(\frac{11.9\text{ mL }min^{-1}}{14.0\text{ mL }min^{-1}})=\sqrt{\frac{32g/mole}{M_1}}$

$M_1=44.2g/mole$

The unknown gas could be carbon dioxide $(CO_2)$ that has approximately 44 g/mole of molar mass.

Thus, the unknown gas could be carbon dioxide $(CO_2)$

5 0

3 years ago

In the reaction a b c, doubling the concentration of a doubles the reaction rate and doubling the concentration of b does not af

PolarNik [594]

The reaction must be a + b --> c

Then you can predict a reaction rate, r o the type r = k * a^n * b^m

Given that the reaction rate is not affected by the concentration of b you can state that m = 0 and r = k * a^n.

Now given, that there is a proportional relation between the reaction rate and a (double a gives double rate), then n = 1 and r = k*a. You can verify that if you dobule a r also doubles.

Answer: r = k*a

3 0

3 years ago