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3 years ago

5

2 C2H6 + 7 026 H2O + 4 CO2

1 answer:

Kay [80]3 years ago

3 0

Answer:

Oxygen is limiting reactant

Explanation:

Based on the chemical reaction:

2C2H6 + 7O2 → 6H2O + 4CO2

<em>2 mole of ethane reacts with 7 moles of oxygen</em>

<em />

For a complete reaction of 5.25 moles of ethane are required:

5.25 moles Ethane * (7mol Oxygen / 2mol Ethane) = 18.38 moles of oxygen

As there are just 15.0 moles of oxygen

<h3>Oxygen is limiting reactant</h3>

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A student does an experiment to determine the molar solubility of lead(II) bromide. She constructs a voltaic cell at 298 K consi

attashe74 [19]

Answer:

The molar solubility of lead bromide at 298K is 0.010 mol/L.

Explanation:

In order to solve this problem, we need to use the Nernst Equaiton:

$E = E^{o} - \frac{0.0591}{n} log\frac{[ox]}{[red]}$

E is the cell potential at a certain instant, E⁰ is the cell potential, n is the number of electrons involved in the redox reaction, [ox] is the concentration of the oxidated specie and [red] is the concentration of the reduced specie.

At equilibrium, E = 0, therefore:

$E^{o} = \frac{0.0591}{n} log \frac{[ox]}{[red]} \\\\log \frac{[ox]}{[red]} = \frac{nE^{o} }{0.0591} \\\\log[red] = log[ox] - \frac{nE^{o} }{0.0591}\\\\[red] = 10^{ log[ox] - \frac{nE^{o} }{0.0591}} \\\\[red] = 10^{ log0.733 - \frac{2x5.45x10^{-2} }{0.0591}}\\\\$

[red] = 0.010 M

The reduction will happen in the anode, therefore, the concentration of the reduced specie is equivalent to the molar solubility of lead bromide.

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3 years ago

How many liters of fat would have to be removed to result in a 5.2 lb weight loss? The density of human fat is 0.94 g/mL?

aksik [14]

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2 years ago

A What does the data tell us? Using the storm data and the computer model, check all statements that are true.

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3 years ago

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1 year ago

Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el

givi [52]

Answer: The amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

Explanation:

We are given:

Moles of electron = 1 mole

According to mole concept:

1 mole of an atom contains $6.022\times 10^{23}$ number of particles.

We know that:

Charge on 1 electron = $1.6\times 10^{-19}C$

Charge on 1 mole of electrons = $1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C$

$Cu^{2+}+2e^-\rightarrow Cu$

$2\times 96500=193000C$ is passed to deposit = 1 mole of copper

63.5 g of copper is deposited by = 193000 C

$14\times 1000g=14000g$ of copper is deposited by = $\frac{193000}{63.5}\times 14000=42551181 C$

To calculate the time required, we use the equation:

$I=\frac{q}{t}$

where,

I = current passed = 40.0 A

q = total charge = 42551181 C

t = time required = ?

Putting values in above equation, we get:

$40.0=\frac{42551181 C}{t}\\\\t=1063779sec$

Converting this into hours, we use the conversion factor:

1 hr = 3600 seconds

So, $1063779s\times \frac{1hr}{3600s}=295hr$

Hence, the amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

3 0

3 years ago