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Semenov [28]
3 years ago
14

6. An environmental chemist needs a carbonate buffer of pH 10.00 to study the effects of acid rain on limestone-rich soils. To p

repare this buffer she will add solid Na2CO3 (106.0 g/mol) to 1.50 L of 0.20 M NaHCO3 . (Ka = 4.7 x 10−11 ) HCO3 − (aq) + H2O(l) ↔ CO3 2− (aq) + H3O + (aq) A. What must the final concentration of Na2CO3 be once mixed with the NaHCO3 solution to obtain a buffer with the correct pH?
Chemistry
1 answer:
pishuonlain [190]3 years ago
5 0

Answer:

[Na₂CO₃] = 0.094M

Explanation:

Based on the reaction:

HCO₃⁻(aq) + H₂O(l) ↔ CO₃²⁻(aq) + H₃O⁺(aq)

It is possible to find pH using Henderson-Hasselbalch formula:

pH = pka + log₁₀ [A⁻] / [HA]

Where [A⁻] is concentration of conjugate base,  [CO₃²⁻] = [Na₂CO₃] and  [HA] is concentration of weak acid, [NaHCO₃] = 0.20M.

pH is desire pH and pKa (<em>10.00</em>) is -log pka = -log 4.7x10⁻¹¹ = <em>10.33</em>

<em />

Replacing these values:

10.00 = 10.33 + log₁₀ [Na₂CO₃] / [0.20]

<em> [Na₂CO₃] = 0.094M</em>

<em />

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Molarity of 0.50 mol sugar in 270 mL of solution.
EastWind [94]
<h3>Answer:</h3>

1.85 M

<h3>Explanation:</h3>

<u>We are given;</u>

  • Number of moles as 0.50 mol
  • Volume of the solution is 270 ml

But, 1000 mL = 1 L

  • Thus, volume of the solution is 0.27 L

We are required to calculate the molarity of the solution;

  • Molarity refers to the concentration of a solution in moles per liter.
  • It is calculated by dividing number of moles with the volume.

Molarity = Moles ÷ Volume

In this case;

Molarity = 0.50 moles ÷ 0.27 L

             = 1.85 Mol/L or 1.85 M

Therefore, molarity of the solution is 1.85 M

5 0
3 years ago
Help mee please (I will give the person who answers this a brainlist)
Leokris [45]
2 5 3 4 1 is the answers
7 0
3 years ago
Read 2 more answers
While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting e
sp2606 [1]

The question is incomplete, here is the complete question:

While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene with water vapor at elevated temperatures.

A chemical engineer studying this reaction fills a 1.5 L flask at 12°C with 1.8 atm of ethylene gas and 4.7 atm of water vapor. When the mixture has come to equilibrium she determines that it contains 1.16 atm of ethylene gas and 4.06 atm of water vapor.

The engineer then adds another 1.2 atm of ethylene, and allows the mixture to come to equilibrium again. Calculate the pressure of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.

<u>Answer:</u> The partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

<u>Explanation:</u>

We are given:

Initial partial pressure of ethylene gas = 1.8 atm

Initial partial pressure of water vapor = 4.7 atm

Equilibrium partial pressure of ethylene gas = 1.16 atm

Equilibrium partial pressure of water vapor = 4.06 atm

The chemical equation for the reaction of ethylene gas and water vapor follows:

                     CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)

<u>Initial:</u>                  1.8                4.7

<u>At eqllm:</u>           1.8-x             4.7-x

Evaluating the value of 'x'

\Rightarrow (1.8-x)=1.16\\\\x=0.64

The expression of K_p for above equation follows:

K_p=\frac{p_{CH_3CH_2OH}}{p_{CH_2CH_2}\times p_{H_2O}}

p_{CH_2CH_2}=1.16atm\\p_{H_2O}=4.06atm\\p_{CH_3CH_2OH}=0.64atm

Putting values in above expression, we get:

K_p=\frac{0.64}{1.16\times 4.06}\\\\K_p=0.136

When more ethylene is added, the equilibrium gets re-established.

Partial pressure of ethylene added = 1.2 atm

                     CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)

<u>Initial:</u>                2.36             4.06               0.64

<u>At eqllm:</u>           2.36-x        4.06-x             0.64+x

Putting value in the equilibrium constant expression, we get:

0.136=\frac{(0.64+x)}{(2.36-x)\times (4.06-x)}\\\\x=0.363,13.41

Neglecting the value of x = 13.41 because equilibrium partial pressure of ethylene and water vapor will become negative, which is not possible.

So, equilibrium partial pressure of ethanol = (0.64 + x) = (0.64 + 0.363) = 1.003 atm

Hence, the partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

3 0
3 years ago
A 0.200 g sample of unknown metal x is dropped into hydrochloride acid and realeases 80.3 mL of hydrogen gas at STP using ideal
s344n2d4d5 [400]

Answer:

The number of mole of the unknown metal is 3.58×10¯³ mole

Explanation:

We'll begin by calculating the number of mole hydrogen gas, H2 that will occupy 80.3 mL at stp.

This is illustrated below:

Recall:

1 mole of any occupy 22.4L or 22400 mL at stp.

1 mole of H2 occupies 22400 mL at stp.

Therefore, Xmol of H2 will occupy 80.3 mL at stp i.e

Xmol of H2 = 80.3/22400

Xmol of H2 = 3.58×10¯³ mole

Therefore, 3.58×10¯³ mole of Hydrogen gas was released.

Now, we can determine the mole of the unknown metal as follow:

The balanced equation for the reaction is given below:

X + 2HCl —> XCl2 + H2

From the balanced equation above,

1 mole of the unknown metal reacted to produce 1 mole of H2.

Therefore, 3.58×10¯³ mole of the unknown metal will also react to produce 3.58×10¯³ mole of H2.

Therefore, the number of mole of the unknown compound is 3.58×10¯³ mole.

5 0
3 years ago
If you mix a 25.0mL sample of a 1.20m potassium sulfide solution with 15.0 mL of a 0.900m basium nitrate solution and the reacti
ddd [48]

Answer:

Limiting reagent: barium nitrate

Theoretical yield: 2.29 g BaS

Percent yield: 87%

Explanation:

The corrected balanced reaction equation is:

K₂S + Ba(NO₃)₂ ⇒ 2KNO₃ + BaS

The amount of potassium sulfide added is:

(25.0 mL)(1.20mol/L) = 30 mmol

The amount of barium nitrate added is:

(15.0mL)(0.900mol/L = 13.5 mmol

Since the reactant react in a 1:1 molar ratio, barium nitrate is the limiting reagent. If all of the limiting reagent reacts, the amount of barium sulfide produced is:

(13.5 mmol Ba(NO₃)₂)(BaS/Ba(NO₃)₂ ) = 13.5 mmol BaS

Converting this amount to grams gives the theoretical yield of BaS (molar mass 169.39 g/mol).

(13.5 mmol)(169.39 g/mol)(1g/1000mg) = 2.29 g BaS

The percent yield is calculated as follows:

(actual yield) / (theoretical yield) x 100%

(2.0 g) / (2.29 g) x 100% = 87%

6 0
3 years ago
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