Answer:
ano poh paki ult kasi hindi mahintindihan yan question mo hindi mahintindihwn
Answer:
case1.
The addition of acid and base leads to a change in pH of the water when adding to deionized water due to fact that acid and bases dissociated in dissolving in water. If the H+ ion increases in the water as acid addition hikes it, it will result in decreasing the pH value. The intensity of the acid also affects the dissociation of the ions.
case2
Buffers are normally formed by weak acid and its conjugate base, and adding acid to the buffer it absorbs the H+ ions so the pH will be lower and adding base or increase of OH- conjugate base resists the pH value to increase.
Nonmetals have the ability to attract electrons better than metals because they have a higher electron affinity or electronegativity than metals.
<h3>What is electronegativity?</h3>
Electronegativity is the tendency, or a measure of the ability, of an atom or molecule to attract electrons and thus form bonds.
An element in the periodic table with a high electronegativity will automatically have a high electron affinity.
Metals (low electronegativity) are known to lose electrons to non-metals (high electronegativity), hence, nonmetals have the ability to attract electrons better than metals because they have a higher electron affinity or electronegativity than metals.
Learn more about electronegativity at: brainly.com/question/2060520
#SPJ1
Answer:
The concentration of H⁺ in a 2.5 M HCl solution is 2.5 M
Explanation:
As HCl is a strong acid and hence a strong electrolyte, it will dissociate as
HCl ⟶ H⁺ + Cl⁻
So, The concentration of H⁺ will be 2.5 M (same as HCl)
Thus, The concentration of H⁺ in a 2.5 M HCl solution is 2.5 M
<u>-TheUnknownScientist</u><u> 72</u>
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:
Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:
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:
Therefore, the leftover of sodium nitrate is:
Finally, the percent yield is computed via:
Best regards!
