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Shkiper50 [21]
3 years ago
7

A buffer is prepared by adding 18.0 g of sodium acetate (ch3coona to 510 ml of a 0.160 m acetic acid (ch3cooh solution. determin

e the buffer
Chemistry
1 answer:
Nutka1998 [239]3 years ago
4 0
Given:

mass of sodium acetate = 18.0 grams
volume of acetic acid = 510 ml
concentration of acetic acid = 0.160 M

Solve for the ph of the buffer:

pH = pKa + log [salt]/[acid]

pKa of acetic acid = 4.76
M of sodium acetate = 82g/mol
pH = 4.76 + log [18g/82g/mol] / [0.160M*(510/1000)]
pH = 5.19


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<u>Answer:</u> The chemical reaction is given below.

<u>Explanation:</u>

A fuel cell is defined as the electrochemical cell which converts the chemical energy of a fuel (often used hydrogen) and an oxidizing agent (often used oxygen) into electrical energy via a pair of redox reactions.

The reactions which occur in hydrogen-oxygen fuel cell are:

At cathode:  H_2+2OH^-\rightarrow 2H_2O+2e^-

At anode:  \frac{1}{2}O_2+H_2O+2e^-\rightarrow 2OH^-

Net reaction:  H_2+\frac{1}{2}O_2\rightarrow H_2O

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4 years ago
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1 C3H8 + 5 O2 --&gt; 3 CO2 + 4H20. If 1.5 moles of C3H8 react, how many
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Answer:

First confirm the reaction is balanced:

C3H8 + 5O2 --> 3CO2 + 4H20 (3 cabon - check; 8 hydrogen - check; 10 oxygen - check).

a) In the equation there is a 5:1 ratio between propane and oxygen.  We also know that number of mole is proportional to pressure and volume.  Since pressure is constant (STP) then the volume of O2 is 7.2 * 5 = 36 litres.

b) For a near ideal gas that PV = nRT (combined gas law).  So for 7.2 litres propane we find n(propane) = 101.3 * 7.2/8.314*298 ~ 0.29 mole (using metric units throughout for simplicity).

There is a 1:3 ratio between propane and CO2.  Therefore 3 * 0.29 = 0.87 mole of CO2 is produced.

MW(CO2) ~ 44 g/mol.  Therefore m(CO2) = 44 * 0.87 ~ 38.3 g

c) We know we need more oxygen than propane (due to the 1:5 ratio) so oxygen is the limiting reagent.  Again Volume is proportional to number of mole and we see there is a 5:4 ratio between oxygen and water.  Therefore the volume of water vapour produced will be (4/5) * 15 = 12 litres.

The other questions use the same technique and will give you some much needed practice.

Explanation:

7 0
4 years ago
How many grams of H2 are needed to produce 13.33g of NH3?
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The reaction to form NH3 is : N2 + 3H2-> 2NH3 12,33g NH3 is 12,33/17,03=0,3 =0,724 moles of NH3 moles NH3. So you need 1,5*0,724 = 1,086 moles H2 1,086*2,016 = 2,189 g of H2 is needed ro form 12,33 g NH3

3 0
4 years ago
A student finds two unlabeled flasks of clear liquids. One is believed to be 0.1 m nacl and the other to be 0.1 m naclo3. What i
KonstantinChe [14]

Answer:

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Explanation:

<u>1. Adding AgNO₃ to NaCl solution:</u>

  • AgNO₃ (aq) + NaCl (aq) → AgCl (s) + NaNO₃ (aq)

<u>2. Adding AgNO₃ to NaClO₃ solution</u>

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<u />

<u>3. Relevant solubility rules for the problem.</u>

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  • All chlorates are soluble, so AgClO₃ is soluble.

  • Salts containing nitrate ion (NO₃⁻) are generally soluble and NaNO₃ is not an exception to this rule. In fact, NaNO₃ is very well known to be soluble.

Hence, when you add AgNO₃ to the NaCl solution the AgCl formed will precipitate, and when you add the same salt (AgNO₃) to the AgClO₃ solution both formed salts AgClO₃ and NaNO₃ are soluble.

Then, the precipiate will permit to conclude which flask contains AgCl.

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3 years ago
Please select the best answer and click "submit."
vladimir1956 [14]
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4 years ago
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