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3 years ago

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If you have ever peeled the label off a glass jar, you may have noticed that the glue does not easily wash off with water. Howev

er, it can be easily removed with another common household solvent– fingernail polish remover (ethel acetate). What does this tell you about the chemical composition of the glue? Explain your answer.

1 answer:

shepuryov [24]3 years ago

8 0

Answer:

It can be removed by acidic chemicals

Explanation:

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How many grams of NH3 can be produced from 12.0g of H2?

RSB [31]

Answer:

Balanced reaction:

3 H2 (g) + N2 (g) → 2 NH3 (g)

Use stoichiometry to convert g of H2 to g of NH3. The process would be:

g H2 → mol H2 → mol NH3 → g NH3

12.0 g H2 x (1 mol H2 / 2.02 g H2) x (2 mol NH3 / 3 mol H2) x (17.03 g NH3 / 1 mol NH3) = 67.4 g NH3

Explanation: See above

Hope this helps, friend.

8 0

2 years ago

You have to prepare 100.0 mL of a 0.100 M solution of sodium carbonate. You have a concentrated solution of sodium carbonate tha

Georgia [21]

Answer:

6.9 ml of concentrate

Explanation:

100 ml of .1 M will require .01 moles

from a 1.45 M solution, .01 mole would be

.01 mole / ( 1.45 mole / liter) = 6.9 ml of the concentrate then dilute to 100 ml

4 0

2 years ago

What particles move most rapidly?

Svet_ta [14]

what are the options?

6 0

3 years ago

Read 2 more answers

25.0cm3 of s saturated potassium hydroxide is neutralized by 35.0cm3 of hydrogen chloride acid of concentration 0.75 mol/dm3. Ca

Kamila [148]

Answer:

Concentration of the original $\rm KOH$ solution: approximately $1.05\; \rm mol \cdot dm^{-3}$ .

Explanation:

Notice that the concentration of the $\rm HCl$ solution is in the unit $\rm mol\cdot dm^{-3}$ . However, the unit of the two volumes is $\rm cm^{3}$ . Convert the unit of the two volumes to $\rm dm^{3}$ to match the unit of concentration.

$\begin{aligned} V(\mathrm{NaOH}) &= 25.0\; \rm cm^{3} \\ &= 25.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0250\; \rm dm^{3} \end{aligned}$ .

$\begin{aligned} V(\mathrm{HCl}) &= 35.0\; \rm cm^{3} \\ &= 35.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0350\; \rm dm^{3} \end{aligned}$ .

Calculate the number of moles of $\rm HCl$ molecules in that $0.0350\; \rm dm^{3}$ of $0.75\; \rm mol \cdot dm^{3}$ $\rm HCl\!$ solution:

$\begin{aligned}n(\mathrm{HCl}) &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.00350\; \rm dm^{3} \times 0.75\; \rm mol \cdot dm^{3} \\ &\approx 0.02625\; \rm mol\end{aligned}$ .

$\rm HCl$ is a monoprotic acid. In other words, each $\rm HCl\!$ would release up to one proton $\rm H^{+}$ .

On the other hand, $\rm KOH$ is a monoprotic base. Each $\rm KOH\!$ formula unit would react with up to one $\rm H^{+}$ .

Hence, $\rm HCl$ molecules and $\rm KOH\!$ formula units would react at a one-to-one ratio.

${\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l)$ .

Therefore, that $0.02625\; \rm mol$ of $\rm HCl$ molecules would neutralize exactly the same number of $\rm NaOH$ formula units. That is: $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ .

Calculate the concentration of a $\rm NaOH$ solution where $V(\mathrm{NaOH}) = 0.0250\; \rm dm^{3}$ and $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ :

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} \\ &= \frac{0.02625\; \rm mol}{0.0250\; \rm dm^{3}}\approx 1.05\; \rm mol \cdot dm^{-3}\end{aligned}$ .

7 0

3 years ago

Given that a mass of a ball is (36.0 g), and the volume is (8.0 mL), what is the ball's density in g/mL?

max2010maxim [7]

Answer:

4.5g/mL

Explanation:

Given parameters:

Mass of ball = 36g

Volume of the ball = 8mL

Unknown:

Density of the ball = ?

Solution:

Density is the mass per unit volume of a substance.

Density = $\frac{mass}{volume}$

So;

Density = $\frac{36}{8}$ = 4.5g/mL

5 0

3 years ago