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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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For the chemical reaction , I identify the reactant and the products .
svetlana [45]

The reactant is Mercury (II) Oxide while the products are Mercury and Oxygen separately.

This is because the reactants are typically always on the left side of the yields symbol. In this decomposition reaction, it would still be the same as at the end of the reaction, there were to products produced: Mercury and Oxygen.

Products tend to always be on the right side of the yields symbol, they're what comes out of a reaction no matter what type.

Hope this helps!

7 0
3 years ago
Yeast and other organisms can convert glucose (C6H12O6) to Ethanol (CH3CH2OH) through a process called Alcoholic Fermentation. T
Murljashka [212]

Answer:

The answer to your question is 8.28 g of glucose

Explanation:

Data

Glucose  (C₆H₁₂O₆) = ?

Ethanol (CH₃CH₂OH)

Carbon dioxide (CO₂) = 2.25 l

Pressure = 1 atm

T = 295°K

Reaction

                             C₆H₁₂O₆    ⇒    2C₂H₅OH(l) +2CO₂(g)

- Calculate the number of moles

                             PV = nRT

Solve for n

                             n = \frac{PV}{RT}

Substitution

                             n = \frac{(1)(2.25)}{(0.082)(295)}

Simplification

                            n = 0.092

- Calculate the mass of glucose

                         1 mol of glucose --------------- 2 moles of carbon dioxide

                          x                         --------------- 0.092 moles

                          x = (0.092 x 1) / 2

                          x = 0.046 moles of glucose

Molecular weight of glucose = 180 g

                       180 g of glucose ---------------  1 mol

                         x  g                    ---------------0.046 moles

                         x = (0.046 x 180) / 1

                         x = 8.28 g of glucose                                    

5 0
3 years ago
What is the mass in grams of<br>1.0mole of (NH4)2S?​
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7 0
3 years ago
The energy of a photon needed to cause ejection of an electron from a photoemissive metal is expressed as the sum of the binding
slega [8]

Answer:

Binding\ energy=43.43\times 10^{-20}\ J

Explanation:

Using the expression for the photoelectric effect as:

E=h\nu_0+\frac {1}{2}\times m\times v^2

Also, E=\frac {h\times c}{\lambda}

\nu_0=\frac {c}{\lambda_0}

Applying the equation as:

\frac {h\times c}{\lambda}=\frac {hc}{\lambda_0}+\frac {1}{2}\times m\times v^2

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

\lambda is the wavelength of the light being bombarded

Given, \lambda=4.00\times 10^{-7}\ m

\frac {hc}{\lambda_0} is the binding energy or threshold energy

\frac {1}{2}\times m\times v^2 is the kinetic energy of the electron emitted.  = 6.26\times 10^{-20}\ J

Thus, applying values as:

\frac{h\times c}{\lambda}=Binding\ Energy+Kinetic\ Energy

\frac{6.626\times 10^{-34}\times 3\times 10^8}{4.00\times 10^{-7}}\ J=Binding\ Energy+6.26\times 10^{-20}\ J

\frac{19.878}{10^{19}\times \:4}\ J=Binding\ Energy+6.26\times 10^{-20}\ J

49.69\times 10^{-20}\ J=Binding\ Energy+6.26\times 10^{-20}\ J

Binding\ energy=43.43\times 10^{-20}\ J

5 0
3 years ago
Is a 93.3% solution dilute or concentrated?<br><br><br> Dilute<br><br><br> Concentrated
oksano4ka [1.4K]

Answer:

<u><em></em></u>

  • <u><em>Concentrated</em></u>

Explanation:

Concentration measures the amount of solute in a solution. There are many expressions of concentration. Some of then are percentage (mass/mass, volume/mass, volume/volume), molarity, molality, mole fraction, among others.

When a solution has a high concentration it is said that it is <em>concentrated; </em>when a solution has a low concentration is is said that is is diluted.

Concentrated solutions expressed in percentage typically have about 80 - 90% (or more) of solute.

Diluted solutions expressed in percentage, tipylcally have about 10% - 20% or less.

But they are not fixed limits. You might say that a 85% solution is concentrated. Acids at 75 % sure are concentrated.

Hence, a 93.3% solution is concentrated, definitely.

3 0
3 years ago
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