All categories

3 years ago

How many liters of O2 at 298 K and 1.00 bar are produced in 1.50 hr in an electrolytic cell operating at a current of 0.0200 A?

Chemistry

1 answer:

stellarik [79]3 years ago

7 0

Answer: 0.0069L

Explanation:

2H2O(l) ---->O2(g) + 4H+(aq) + 4e-

no of moles= it/eF

NO of moles of O2 produced = (Current in Ampere x Time in second)/ (Faraday constant x Number of electrons required)

Moles of O2 produced = (0.02x (60 x 60X1.5 s)/(96485 x 4)

= 0.0002798 moles= 2.798x 10 ^-4moles

Using ideal gas equation,

P V = n R T

Where, P is the pressure,

V is the volume,

n is the number of moles,

R is the gas constant, and T is the temperature

We have, 1 bar = 0.986923 atm

Substituting the values,

V = nRT/P = (2.798 x 10-4moles x 0.08205 L atm mol K x 298 K)/ 0.986923 atm = 0.0069L

Volume of O2 produced = 0.0069L

You might be interested in

You measure the mass of a model car to be 230 grams.The actual mass is 218 grams .what is your percent error

LenKa [72]

Your percent is 5.50458716

8 0

4 years ago

Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe

leva [86]

Answer:

0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as molarity, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

$c = \frac{n}{V}$

n = moles (unit mol)

V = volume (unit L)

Find the molar mass (M) of potassium hydroxide.

$M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}$

$M_{KOH} = 56.106 \frac{g}{mol}$

Calculate the moles of potassium hydroxide.

$n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}$

$n_{KOH} = 0.25941(9)mol$

Carry one insignificant figure (shown in brackets).

Convert the volume of water to litres.

$V = \frac{500.0mL}{1}*\frac{1L}{1000mL}$

$V = 0.5000L$

Here, carrying an insignificant figure doesn't change the value.

Calculate the concentration.

$c = \frac{n}{V}$

$c = \frac{0.25941(9)mol}{0.5000 L}$

$c = 0.5188(3) \frac{mol}{L}$ <= Keep an insignificant figure for rounding

$c = 0.5188 \frac{mol}{L}$ <= Rounded up

$c = 0.5188M$ <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

7 0

3 years ago

If gatorade g series, endurance formula, contains 14 grams of carbohydrate per 8 ounces, how many ounces should a basketball pla

Mashcka [7]

Based on recommended amount of carbohydrate, a basketball player should consume about 17 - 34 ounces of gatorade g series during the hour-long game.

<h3>How many ounces of endurance formula gatorade g series, endurance formula should a basketball player consume during an hour-long game if it contains 14 grams of carbohydrate per 8 ounces?</h3>

Carbohydrates are food substances metabolized easily by the body to produce energy.

Given that the recommended amount of carbohydrate to consume to maintain performance is 30–60 g/h.

Also 14 grams of carbohydrate found in 8 ounces of the drink.

30 g of carbohydrate will be present in 30 × 8/14 = 17.1 ounces of gatorade g series

60 g of carbohydrate will be present in 60 × 8/14 =34.3 ounces of gatorade g series.

Therefore, a basketball player should consume about 17 - 34 ounces of gatorade g series during the hour-long game.

Learn more about carbohydrates at: brainly.com/question/797978

8 0

3 years ago

Identify the element in C6H12O6 that allows it to be classified as an organic compound.

LenaWriter [7]

Answer:

i dont know lol

Explanation:

8 0

3 years ago

Which oh these statements describes a chemical property?

Gelneren [198K]

Iron rusts when exposed to air → chemical property

3 0

3 years ago