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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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A chemist reacted 11.50 grams of sodium metal with an excess amount of chlorine gas. The chemical reaction that occurred is show
iren [92.7K]

Actual yield is the definite mass of the product obtained from the reaction. The actual yield of the reaction is 24.83 gms.

<h3>What is the percentage yield?</h3>

Percentage yield is the proportion of the actual to the theoretical yield. The formula is given as,

% yield = Actual yield ÷ Theoretical yield × 100

The balanced chemical reaction is given as:

2 Na + Cl₂ → 2NaCl

From the reaction:

2 × 23 Na = 2 × 58.44 gm of NaCl

If, 1 gm of Na will produce = (2 × 58.44) ÷ (2 × 23)

Then, 11.50 gms sodium will produce, (2 × 58.44) ÷ (2× 23) × 11.50

= 29.22 g

So, the actual yield is calculated as:

= % yield × Theoretical yield ÷ 100

= 85 × 29.22 ÷ 100

= 24.83 g

Therefore, the actual yield of the reaction is 24.83 gms.

Learn more about actual yield here:

brainly.com/question/22257659

#SPJ1

8 0
3 years ago
One of the significant achievements of Fahrenheit was to: O make thermometers smaller using air O make thermometers compact usin
irga5000 [103]

Answer:

make thermometers smaller using mercury

Explanation:

Daniel Fahrenheit invented first accurate thermometer which used mercury instead of the alcohol and the water mixtures. In laboratory, he used this invention of him to develop first temperature scale which was enough precise to become the worldwide standard.

The key to the Fahrenheit's thermometer was that the mercury is able to rise and fall within tube without sticking to sides. It was ideal substance for the reading temperatures since mercury expanded at more constant rate than the alcohol and is also able to be read the temperature at much higher and also lower temperatures.

3 0
4 years ago
Covalent bonds in a molecule absorb radiation in the IR region and vibrate at characteristic frequencies.
topjm [15]

Answer:

(a) ν = 3.1 × 10¹³ s⁻¹

(b) λ = 3.467 μm

Explanation:

We can solve both problems using the following expression.

c = λ × ν

where,

c: speed of light

λ: wavelength

ν: frequency

(a)

c = λ × ν

ν = c / λ

ν = (3.000 × 10⁸ m/s) / (9.6 × 10⁻⁶ m)

ν = 3.1 × 10¹³ s⁻¹

(b)

c = λ × ν

λ = c / ν

λ = (3.000 × 10⁸ m/s) / (8.652 × 10¹³ s⁻¹)

λ = 3.467 × 10⁻⁶ m

λ = 3.467 × 10⁻⁶ m (10⁶ μm/ 1 m)

λ = 3.467 μm

4 0
3 years ago
How many moles of H2O are produced when 64.0 g C2H2 burns in oxygen
Gnoma [55]
The balanced chemical reaction will be: 

C2H2 + 5/2O2 = 2CO2 + H2O

We are given the amount of C2H2 being burned. This will be our starting point.

64.0 g C2H2 (1 mol C2H2 / 26.02 g C2H2) (1 mol H2O/1mol C2H2) ( 18.02 g H2O/1mol H2O) = 44.32 g H2O.

Thus, the answer is 44.32 g H2O.
6 0
3 years ago
The following shows the precipitation reaction of barium chloride (BaCl₂) and sodium hydroxide (NaOH):
Nadya [2.5K]

Answer:

The answer to your question is:

a) BaCl2

b) 0.8208 g

c) yield = 85.3 %

d)

Explanation:

                     BaCl₂(aq) + NaOH(aq) ----> Ba(OH)₂(s) + 2NaCl(aq)

Data

a) 1 g of BaCl₂

   1 g of NaOH

MW BaCl2 = 137 + (35.5x2) = 208 g

MW NaOH = 23 + 16 + 1 = 40 g

                               208 g of BaCl2 -------------  1 mol

                                    1 g of BaCl2 -------------    x

                                  x = ( 1 x 1) / 208 = 0.0048 mol of BaCl2

                                    40 g of NaOH ------------  1 mol

                                       1 g of NaOH ------------   x

                                 x = (1 x 1) / 40

                                 x = 0.025 mol of NaOH

The ratio BaCl2 to NaOH is 1:1 (in the equation)

But experimentally we have 0.0048 : 0.025, so the limiting reactant is BaCl2, because is in lower concentration.

b)

                    1 mol of BaCl2 -------------- 1 mol of Ba(OH)2

                    0.0048 mol     ---------------   x

                     x = (0.0048 x 1) / 1

                     x = 0.0048 mol of Ba(OH)2

MW Ba(OH)2 = 137 + 32 + 2 = 171 g

                     171 g of Ba(OH)2 -------------------- 1 mol

                      x                         --------------------  0.0048 mol

                     x = (0.0048 x 171) / 1

                     x = 0.8208 g

c)Data

Ba(OH)2 = 0.700 g

                   % yield = 0.700 / 0.8208 x 100

                   % yield = 85.3

d)

Sorry, i don't understand this question

       

6 0
3 years ago
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