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2 years ago

Can I mix a sunlight body wash soap and a lemon to get rid of dark spots?

Chemistry

1 answer:

yuradex [85]2 years ago

6 0

Answer:

Mix ½ teaspoon apple cider vinegar, ½ teaspoon lemon juice, and 1 tablespoon water, and dip a cotton ball in the liquid. Apply to dark

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Calculate the feed ratio of adipic acid and hexamethylene diamine that should be employed to obtain a polyamide of approximately

artcher [175]

Answer:

r= 0.9949 (For 15,000)

r=0.995 (For 19,000)

Explanation:

We know that

Molecular weight of hexamethylene diamine = 116.21 g/mol

Molecular weight of adipic acid = 146.14 g/mol

Molecular weight of water = 18.016 g/mol

As we know that when adipic acid and hexamethylene diamine react then nylon 6, 6 comes out as the final product and release 2 molecule of water.

$M_{repeat}=146.14+166.21-2\times 18.106\ g/mol$

$M_{repeat}=226.32\ g/mol$

Mo= 226.32/2 =113.16 g/mol

$M_n=X_nM_o$

Given that

Mn= 15,000 g/mol

15,000 = Xn x 113.16

Xn = 132.55

Now by using Carothers equation we know that

$X_n=\dfrac{1+r}{1+r-2rp}$

$132.55=\dfrac{1+r}{1+r-2\times 0.99r}$

By calculating we get

r= 0.9949

For 19,000

19,000 = Xn x 113.16

Xn = 167.99

By calculating in same process given above we get

r=0.995

3 0

3 years ago

A 9.79 mol sample of freon gas was placed in a balloon. Adding 3.50 mol of freon gas to the balloon increased its volume to 21.8

aivan3 [116]

Answer:

16.06 L was the initial volume of the balloon.

Explanation:

Initial moles of freon in ballon = $n_1=9.79 mol$

Initial volume of freon gas in ballon = $V_1=?$

Moles of freon gas added in the balloon = n = 3.50 mole

Final moles of freon in ballon = $n_2=n_1+n=9.79 mol+3.50 mol=13.29 mol$

Final volume of freon gas in ballon = $V_2=21.8 L$

Using Avogadro's law:

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$ ( at constant pressure and temperature)

$V_1=\frac{V_2\times n_1}{n_2}=\frac{21.8 L\times 9.79 mol}{13.29 mol}=16.06L$

16.06 L was the initial volume of the balloon.

4 0

2 years ago

20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the

kirill115 [55]

Answer:

Choice A: approximately $0.12\; \rm M$ .

Explanation:

Note that the unit of concentration, $\rm M$ , typically refers to moles per liter (that is: $1\; \rm M = 1\; \rm mol\cdot L^{-1}$ .)

On the other hand, the volume of the two solutions in this question are apparently given in $\rm cm^3$ , which is the same as $\rm mL$ (that is: $1\; \rm cm^{3} = 1\; \rm mL$ .) Convert the unit of volume to liters:

$V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L$ .
$V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L$ .

Calculate the number of moles of $\rm H_2SO_4$ formula units in that $0.02\; \rm L$ of the $0.09\; \rm M$ solution:

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}$ .

Note that $\rm H_2SO_4$ (sulfuric acid) is a diprotic acid. When one mole of $\rm H_2SO_4$ completely dissolves in water, two moles of $\rm H^{+}$ ions will be released.

On the other hand, $\rm NaOH$ (sodium hydroxide) is a monoprotic base. When one mole of $\rm NaOH$ formula units completely dissolve in water, only one mole of $\rm OH^{-}$ ions will be released.

$\rm H^{+}$ ions and $\rm OH^{-}$ ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid $\rm H_2SO_4$ dissolves in water completely, it will take two moles of $\rm OH^{-}$ to neutralize that two moles of $\rm H^{+}$ produced. On the other hand, two moles formula units of the monoprotic base $\rm NaOH$ will be required to produce that two moles of $\rm OH^{-}$ . Therefore, $\rm NaOH$ and $\rm H_2SO_4$ formula units would neutralize each other at a two-to-one ratio.

$\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O$ .

$\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2$ .

Previous calculations show that $0.0018\; \rm mol$ of $\rm H_2SO_4$ was produced. Calculate the number of moles of $\rm NaOH$ formula units required to neutralize that

$\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}$ .

Calculate the concentration of a $0.03\; \rm L$ solution that contains exactly $0.0036\; \rm mol$ of $\rm NaOH$ formula units:

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}$ .

3 0

3 years ago

Consider the following chemical equilibrium:

jok3333 [9.3K]

Answer:

Kc = [CH₄] / [H₂]²

Kp = [CH₄] / [H₂]² * (0.082*T)^-1

Explanation:

Equilibrium constant, Kc, is defined as the ratio of the concentrations of the products over the reactants. Also, each concentration of product of reactant is powered to its coefficient.

<em>Pure solids and liquids are not taken into account in an equilibrium</em>

Thus, for the reaction:

C(s)+ 2H₂(g) ⇌ CH₄(g)

Equilibrium constant is:

Now, using the formula:

Kp = Kc* (RT)^Δn

<em>Where R is gas constant (0.082atmL/molK), T is the temperature of the reaction and Δn is difference in coefficients of gas products - coefficients of gas reactants (1 - 2= -1)</em>

Replacing:

<em />

4 0

3 years ago

What element would have this electron configuration:<br> 1s22s22p 3s 3p 4s 3d

kirill [66]

Answer: its either argon or phosphorus is this a constructed response or a multiple choice

Explanation:

3 0

3 years ago

Can I mix a sunlight body wash soap and a lemon to get rid of dark spots?​

Can I mix a sunlight body wash soap and a lemon to get rid of dark spots?