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Margarita [4]
3 years ago
12

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.
Chemistry
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
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