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Fantom [35]
3 years ago
7

Why is mendeleevâs periodic table considered a viable scientific model?

Chemistry
1 answer:
ludmilkaskok [199]3 years ago
4 0
It could be used to explain the behavior of all elements.
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If you go into the lab and dissolve 2.11 g of magnesium chloride in enough water to make 1500 ml of solution, what is the molari
dusya [7]
Moles pf mgcl2= 2,11/M mgcl2=2,11/95= 0,0222

Molarity=0,0222/1,5=0,0148 M.


I hope this is correct.
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What happens to ice as it warms up
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Ice starts to melt and turns into water as it melts
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Which of the following has the strongest buffering capacity? A. H2O B. 0.1 M HCl C. 0.1 M carbonic/bicarbonate (H2CO3/HCO3-) at
enyata [817]

Explanation:

(A)   As we know that carbonic acid (H_{2}CO_{3}) and Sodium bicarbonate (NaHCO_{3}) forms an acidic buffer.

Therefore, pH of an acidic buffer is given by Hendeerson-Hasselbalch equation as follows.

               pH = pK_{a} + log(\frac{[Salt]}{[Acid]}) ........... (1)

So mathematically,  if [Salt] = [Acid]  then \frac{[Salt]}{[Acid]} = 1 .

And,  log (\frac{[Salt]}{[Acid]}) = 0

Therefore, equation (1) gives us the following.

         pH = pK_{a} (when acid and salt are equal in concentration)

Hence, pK_{a} of H_{2}CO_{3} (carbonic acid) is 6.35.

And, with this we have following results.

In (A) and (D) we have the case \frac{[NaHCO_{3}]}{[H_{2}CO_{3}]}[/tex] i.e. [Salt] = [Acid].

Hence, for the cases pH = pK_{a} = 6.35.

(B)    [NaHCO_{3}] = 0.045 M and,  [H_{2}CO_{3}] = 0.45 M

Hence,   pH = 6.35 + log([NaHCO_{3}][[H_{2}CO_{3}])

                     = 6.35 + log(\frac{0.045}{0.45})

                     = 6.35 + (-1)

                     = 5.35

Therefore, it means that this buffer will be most suitable buffer as it has pH on acidic side and addition of slight excess base will not affect much of its pH value.

(C)    [NaHCO_{3}] = 0.45 M [H_{2}CO_{3}]

                          = 0.045 M

So,       pH = 6.35 + log(\frac{[NaHCO_{3}]}{[H_{2}CO_{3}]})

                  = 6.35 + log(\frac{0.45}{0.045})

                  = 6.35 + (+1)

                 = 7.35

This means that pH on Basic side makes it no more acidic buffer.

5 0
3 years ago
Determine the resulting pH when 75 mL of 0.05M HBr is reacted with 74 mL of 0.075 M KOH
Sunny_sXe [5.5K]

Answer:

pH = 12.08

Explanation:

  • H⁺ + OH⁻ → H₂O

First we <u>calculate how many moles of each substance were added</u>, using <em>the given volume and concentration</em>:

  • HBr ⇒ 0.05 M * 75 mL = 3.75 mmol HBr
  • KOH ⇒ 0.075 M * 74 mL = 5.55 mmol KOH

As HBr is a strong acid, it dissociates completely into H⁺ and Br⁻ species. Conversely, KOH dissociates completely into OH⁻ and K⁺ species.

As there are more OH⁻ moles than H⁺ moles (5.55 vs 3.75), we <u>calculate how many OH⁻ moles remain after the reaction</u>:

  • 5.55 - 3.75 = 1.8 mmoles OH⁻

With that<em> number of moles and the volume of the mixture</em>, we <u>calculate [OH⁻]</u>:

  • [OH⁻] = 1.8 mmol / (75 mL + 74 mL) = 0.0121 M

With [OH⁻], we <u>calculate the pOH</u>:

  • pOH = -log[OH⁻] = 1.92

With the pOH, we <u>calculate the pH</u>:

  • pH = 14 - pOH
  • pH = 12.08
6 0
3 years ago
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