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notsponge [240]
1 year ago
7

Consider the cell notation Cd(s) | Cd(NO3)2(aq) || AgNO3(aq) | Ag(s) with standard reduction potentials for Cd2+ and Ag+ as -0.4

02V and 0.799V respectively. What conclusions would you make for the reaction after determining the Gibbs free energy (△G)
Chemistry
1 answer:
MakcuM [25]1 year ago
8 0

The value of the Gibbs free energy shows us that the reaction is spontaneous.

<h3>What is the Gibbs free energy?</h3>

The Gibbs free energy is a quantity that helps us to be able to determine the spontaneity of a reaction.

In order to obtain the Gibbs free energy, we must obtain the Ecell as follows; 0.799V - (-0.402V) = 1.201 V

Now;

△G = -nFEcell

△G = -(2 * 96500 * 1.201)

△G = -232kJ/mol

Learn more about free energy:brainly.com/question/15319033?

#SPJ1

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6.75 L of carbon dioxide gas are stored at a pressure of 175 atm. If the temperature and number of gas particles do not change,
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Answer:

4.24 L

Explanation:

  • 6.75 L - 175 atm

x. - 110 atm

  • <u>175atm </u><u>x</u>=<u>742.5L /atm</u>

175atm 175atm

  • therefore x=4.24 L
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What is the boiling point of water at sea level ??​
TEA [102]

Answer:

212 °F

Explanation:

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Atomic size of inert gases do not affect you inertness why​
Tresset [83]

Answer:

Because your body has built-in resistance to certain gases, no matter the size of the gas cloud.

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What are the names of the stable forms of oxygen?
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2 years ago
A chemistry graduate student is given 125.mL of a 1.00M benzoic acid HC6H5CO2 solution. Benzoic acid is a weak acid with =Ka×6.3
lubasha [3.4K]

Answer:

53.9 g

Explanation:

When talking about buffers is very common the problem involves the use of the Henderson Hasselbach formula:

pH = pKa + log [A⁻]/[HA]

where  [A⁻] is the concentration of the conjugate base of the weak acid HA, and [HA] is the concentration of the weak acid.

We can calculate pKₐ from the given kₐ ( pKₐ = - log Kₐ ), and from there obtain the ratio  [A⁻]/HA].

Since we know the concentration of HC6H5CO2 and the volume of solution, the moles and mass of KC6H5CO2  can be determined.

So,

4.63 = - log ( 6.3 x 10⁻⁵ ) + log [A⁻]/[HA] = - (-4.20 ) + log [A⁻]/[HA]

⇒ log [A⁻]/[HA]  = 4.63 - 4.20 =  log [A⁻]/[HA]

0.43 = log [A⁻]/[HA]

taking antilogs to both sides of this equation:

10^0.43 =  [A⁻]/[HA] = 2.69

 [A⁻]/ 1.00 M = 2.69 ⇒ [A⁻] = 2.69 M

Molarity is moles per liter of solution, so we can calculate how many moles of  C6H5CO2⁻ the student needs to dissolve  in 125. mL ( 0.125 L ) of a 2.69 M solution:

( 2.69 mol C6H5CO2⁻ / 1L ) x 0.125 L  = 0.34 mol C6H5CO2⁻

The mass will be obtained by multiplying 0.34 mol times molecular weight for KC6H5CO2 ( 160.21 g/mol ):

0.34 mol x 160.21 g/mol = 53.9 g

3 0
3 years ago
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