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3 years ago

Do bases contain hydrogen ions (H+)

1 answer:

6 0

One definition of an acid is a molecule that can donate a H+ ion. So for example if you have an acid such as hydrochloric acid (HCl) in water, HCl will donate it's H+ to the H2O molecule, forming Cl- and H3O+.

In contrast, a base is a molecule that accepts H+ ions. This means a base such as NH3 reacts with H2O to form OH- and NH4.

So when pH is measured, it is the ratio of H3O+ ions to OH- ions. If there are more H3O+ ions, you then know that there are more acidic molecules in the solution, since they must be giving away their H+ ions.

So a solution with an acid in it will have more H3O+ ions (which can be described as H+ ions) in it. However, an acid by itself will have nothing to donate H+ ions to.

hope this helps also please make me the brainliest

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2 years ago

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!

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2 years ago