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choli [55]
3 years ago
15

A 250 ml solution of 2.0 M NaOH is diluted to 1.0 liter. What is the final concentration of the solution?

Chemistry
2 answers:
Alex73 [517]3 years ago
3 0

Answer:

Diluted concentration is 0.5M

Explanation:

Let's solve this with rules of three, although there is a formula to see it easier

In 1000 mL (1L), we have 2 moles of NaOH

In 250 mL we must have (250 . 2) / 1000 = 0.5 moles of NaOH

These moles will be also in 1 L of the final volume of the diluted solution

More easy:

1 L of solution has 0.5 moles of NaOH

Then, molarity is 0.5 M

The formula is: Concentrated M . Conc. volume = Diluted M . Diluted volume

2 M . 0.250L = 1L . Diluted M

0.5M = Diluted M

Leto [7]3 years ago
3 0

Answer:

The final concentration of the solution is 0.500 M

Explanation:

Step 1: data given

Initial volume = 250 mL = 0.250 L

Molarity NaOH = 2.0 M

The volume is diluted to 1.0 L

Step 2: Calculate the final concentration of the solution

C1*V1 = C2*V2

⇒with C1 = the initial concentration of NaOH = 2.0 M

⇒with V1 = the initial volume = 250 mL = 0.250 L

⇒with C2 = the final concentration of the solution = TO BE DETERMINED

⇒with v2 = the volume of the diluted solution = 1.0 L

2.0 M * 0.250 L = C2 * 1.0 L

C2 = 2.0* 0.250

C2 = 0.500 M

The final concentration of the solution is 0.500 M

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500 mL of water is added to 400 mL of 0.35 M HCl. Find the concentration of the diluted solution.
jarptica [38.1K]

The concentration of diluted solution is 0.16 M

<u>Explanation:</u>

As, the number of moles of diluted solution and concentrated solution will be same.

So, the equation used to calculate concentration will be:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the concentrated HCl solution

M_2\text{ and }V_2 are the molarity and volume of diluted HCl solution

We are given:

M_1=0.35M\\V_1=400mL\\M_2=?M\\V_2=(500+400)mL=900mL

Putting values in above equation, we get:

0.35\times 400=M_2\times 900\\\\M_2=\frac{0.35\times 400}{900}=0.16M

Hence, the concentration of diluted solution is 0.16 M

5 0
3 years ago
Which term describes the molecule shown below h-c-c-h ethyne c2h2?
Andru [333]
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4 0
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The element gallium has an atomic weight of 69.7 and consists of two stable isotopes gallium -69 and gallium -71. The isotope ga
BigorU [14]

Answer:

The answer to your question is Gallium-71 = 70.9202 amu

Explanation:

Gallium atomic weight = 69.7

Gallium-69 = 68.9 amu    abundance = 60.4%

Gallium-71 =     x                abundance = 39.6%  

To solve this problem just write an equation and solve it for the mass of gallium-71.

Equation

Gallium = Gallium-69(abundance) + Gallium-71(abundance)

Substitution

69.7 = (68.9)(0.604) + Gallium-71(0.396)

69.7 = 41.6156 + Gallium-71(0.396)

Gallium-71(0.396) = 69.7 - 41.6156

Gallium-71(0.396) = 28.0844

Gallium-71 = 28.0844/0.396

Gallium-71 = 70.9202 amu

3 0
2 years ago
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
Why the statement all atoms of one element have the same mass is false
White raven [17]
Some elements have isotopes which have a different number of neutrons, and this means they have different masses.
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