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Annette [7]
3 years ago
11

A buffer solution contains 0.368 M hydrocyanic acid and 0.360 M potassium cyanide . If 0.0513 moles of sodium hydroxide are adde

d to 225 mL of this buffer, what is the pH of the resulting solution
Chemistry
1 answer:
MrRa [10]3 years ago
7 0

Solution :

Millimoles of hydrocyanic acid = 225 x 0.368

                                                   = 82.8

Millimoles of potassium cyanide = 225 x 0.360

                                                   = 81

Millimoles of sodium hydroxide = 51.3

Therefore,

pOH = pKb + log [salt - C / bas + C]

        = 4.74 + log[82.8 - 51.3 / 81 + 51.3]

         = 4.102

Therefore, pH = 9.05

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Explain how the cardiovascular system helps your body function
zhuklara [117]

Answer:

    The blood circulatory system (cardiovascular system) delivers nutrients and oxygen to all cells in the body. It consists of the heart and the blood vessels running through the entire body. The arteries carry blood away from the heart; the veins carry it back to the heart.

Explanation:

You're welcome!

Stay Safe!

4 0
2 years ago
You need to prepare 100.0 mL of a pH 4.00 buffer solution using 0.100 M benzoic acid (p K a = 4.20 ) and 0.200 M sodium benzoate
bekas [8.4K]

Answer : The volume of sodium benzoate and benzoic acid  solution mixed to prepare this buffer should be, 29.0 mL and 71 mL respectively.

Explanation :

Let the volume of sodium benzoate (salt) be, x

So, the volume of benzoic acid  (acid) will be, (100 - x)

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

Now put all the given values in this expression, we get:

4.00=4.20+\log \left(\frac{(\frac{0.200x}{100})}{(\frac{0.100(100-x)}{100})}\right)

x = 29.0

The volume of sodium benzoate = x = 29.0 mL

The volume of benzoic acid  (acid) = (100 - x) = (100 - 29.0) = 71 mL

Thus, the volume of sodium benzoate and benzoic acid  solution mixed to prepare this buffer should be, 29.0 mL and 71 mL respectively.

5 0
3 years ago
A 0.450 g sample of solid lead(II) nitrate is added to 250 mL of 0.250 M sodium iodide solution. Assume no change in volume of t
Verdich [7]

Pb(NO₃)₂ ⇒limiting reactant

moles PbI₂ = 1.36 x 10⁻³

% yield  = 87.72%

<h3>Further explanation</h3>

Given

Reaction(unbalanced)

Pb(NO₃)₂(s) + NaI(aq) → PbI₂(s) + NaNO₃(aq)

Required

  • moles of PbI₂
  • Limiting reactant
  • % yield

Solution

Balanced equation :

Pb(NO₃)₂(s) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)

mol Pb(NO₃)₂ :

= 0.45 : 331 g/mol

= 1.36 x 10⁻³

mol NaI :

= 250 ml x 0.25 M

= 0.0625

Limiting reactant (mol : coefficient)

Pb(NO₃)₂ : 1.36 x 10⁻³ : 1 = 1.36 x 10⁻³

NaI : 0.0625 : 2 = 0.03125

Pb(NO₃)₂ ⇒limiting reactant(smaller ratio)

moles PbI₂ = moles Pb(NO₃)₂ = 1.36 x 10⁻³(mol ratio 1 : 1)

Mass of PbI₂ :

= mol x MW

=  1.36 x 10⁻³ x 461,01 g/mol

= 0.627 g

% yield = 0.55/0.627 x 100% = 87.72%

7 0
3 years ago
Area A receives more than ten inches of rain per year Area B receives less than one inch of rain per year Which of the following
STALIN [3.7K]

Answer:

Area A has a lot more annual rainfall than area B.

Explanation:

If area A has more than 10 in. rain, and area B has less than 1 in. annual, then area A will obviously get more rainfall. Hope this helps!

3 0
3 years ago
Properties of a longitudinal wave
AleksAgata [21]
In longitudinal waves, the oscillations are along the same direction as the direction of travel and energy transfer.
Sound waves and waves in a stretched spring are longitudinal waves. P waves (relatively fast moving longitudinal seismic waves that travel through liquids and solids) are also longitudinal waves.
Longitudinal waves show area of compression and rarefaction. In the animation, the areas of compression are where the parts of the spring are close together, while the areas of rarefaction are where they are far apart.
8 0
3 years ago
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