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Arturiano [62]
2 years ago
9

Where would hcl lie on the pH scale?

Chemistry
1 answer:
Valentin [98]2 years ago
6 0
It is very low on the ph scale because it has a 2 ph level.
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Stemscopedia
Alexeev081 [22]
What’s the question
7 0
3 years ago
Select the correct answer.
zalisa [80]

Well, this is somewhat difficult, because an electron already creates an electric field.  However, I know that when an electron moves it then creates a magnetic field.  So, I'm going to safely assume that when an electron moves, it creates an electric, a magnetic, and a gravitational field.  

I hope that helps!

8 0
3 years ago
Read 2 more answers
If 15.6 grams of copper (ii) chloride react with 20.2 grams of sodium nitrate how many grams of sodium chloride can be formed? W
olasank [31]

Answer:

- 13.56 g of sodium chloride are theoretically yielded.

- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.

- 0.50 g of sodium nitrate remain when the reaction stops.

- 92.9 % is the percent yield.

Explanation:

Hello!

In this case, according to the question, it is possible to set up the following chemical reaction:

CuCl_2+2NaNO_3\rightarrow 2NaCl+Cu(NO_3)_2

Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:

m_{NaCl}^{by\ CuCl_2}=15.6gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2} *\frac{2molNaCl}{1molCuCl_2} *\frac{58.44gNaCl}{1molNaCl} =13.56gNaCl\\\\m_{NaCl}^{by\ NaNO_3}=20.2gNaNO_3*\frac{1molNaNO_3}{84.99gNaNO_3} *\frac{2molNaCl}{2molNaNO_3} *\frac{58.44gNaCl}{1molNaCl} =13.89gNaCl

Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:

m_{NaNO_3}^{by\ NaCl}=13.56gNaCl*\frac{1molNaCl}{58.44gNaCl}*\frac{2molNaNO_3}{2molNaCl} *\frac{84.99gNaNO_3}{1molNaNO_3}=19.72gNaNO_3

Therefore, the leftover of sodium nitrate is:

m_{NaNO_3}^{leftover}=20.2g-19.7g=0.5gNaNO_3

Finally, the percent yield is computed via:

Y=\frac{12.6g}{13.56g} *100\%\\\\Y=92.9\%

Best regards!

6 0
2 years ago
Could the structures below undergo a Fischer esterification reaction? Reaction scheme of benzoic acid and thionyl chloride to fo
Romashka-Z-Leto [24]

Answer:

The correct answer is the first option. No, the structures above cannot undergo a Fischer esterification reaction to form an ester.

Explanation:

The reaction that will take place can be found in the attached file. The reaction does not require any catalyst and it cannot undergo a Fischer esterification reaction to form an ester.

4 0
2 years ago
Which equation using element symbols correctly describes the reaction: "One molecule of methane plus two molecules of diatomic o
MrMuchimi

Answer:

CH4+2O2→2H2O+CO2

8 0
3 years ago
Read 2 more answers
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