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s2008m [1.1K]
3 years ago
5

Calculate the standard potential, e°, for this reaction from its equilibrium constant at 298 k. 6.47 x 10^5

Chemistry
1 answer:
Elina [12.6K]3 years ago
4 0
The working equation for this is written below:

E° = 0.0592logK/n

So, we have to know the value of n first which represents the number of moles electron in the reaction. We cannot answer this because we are not given with the reaction. However, just suppose the reaction is:

Cu⁺ + 2e⁻ --> Cu

Then, n=2. Continuing,

E° = 0.0592log(6.47×10⁵)/2 =<em> 0.172 V</em>
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Which of the following substances would likely dissolve but not dissociate in water
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Oils, fats, and certain organic solvents do not dissolve in water because they are hydrophobic.

Explanation:

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2 years ago
What is the pressure in a 5.00 L tank with 49.00 grams of oxygen gas at 350 K? ___atm
ziro4ka [17]

Answer:

The right answer is "8.81 atm".

Explanation:

Given:

V = 5.00 L

Mass = 4900 g

MW = 32 g/mol

T = 350 K

Now,

Number of moles will be:

n = \frac{Mass}{MW}

   =\frac{49.00}{32}

   =1.53125 \ mole

By using the ideal gas equation, we get

⇒ PV=nRT

or,

⇒    P=\frac{nRT}{V}

By substituting the values, we get

           =\frac{1.53125\times 0.0821\times 350}{5.00}

           =\frac{44.1}{5.00}

           =8.81 \ atm  

4 0
3 years ago
Soils that allow water to pass through them faster are more _______.
siniylev [52]
I would say C. Permable

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4 0
3 years ago
Read 2 more answers
If a 1.00 mL sample of the reaction mixture for the equilibrium constant experiment required 32.40 mL of 0.258 M NaOH to titrate
andrey2020 [161]

Answer:

The concentration of acetic acid is 8.36 M

Explanation:

Step 1: Data given

Volume of acetic acid = 1.00 mL = 0.001 L

Volume of NaOH = 32.40 mL = 0.03240 L

Molarity of NaOH = 0.258 M

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate the concentration of the acetic acid

b*Ca*Va = a*Cb*Vb

⇒with b = the coefficient of NaOH = 1

⇒with Ca = the concentration of CH3COOH = TO BE DETERMINED

⇒with Va = the volume of CH3COOH = 1.00 mL = 0.001L

⇒with a = the coefficient of CH3COOH = 1

⇒with Cb = the concentration of NaOH = 0.258 M

⇒with Vb = the volume of NaOH = 32.40 mL = 0.03240 L

Ca * 0.001 L = 0.258 * 0.03240

Ca = 8.36 M

The concentration of acetic acid is 8.36 M

6 0
3 years ago
6. Which example has BOTH ionic and covalent bonds? *
Fed [463]

Answer:

It’s either A or D  buttttt probably d

Explanation:

7 0
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