A salt solution with a volume of 250 mL that contains 0.70 mol NaCl.

The equation shows one mole of ethanol fuel being burned in oxygen. Convert the energy released into its equivalent mass. C2H5OH

lana [24]

Answer:

47.9 g of ethanol

Explanation:

Combustion is a chemical reaction in which a substance reacts with oxygen to produce heat and light. Combustion reactions have been very useful as a source of energy. Ethanol is now burnt for energy purposes as a fuel. Ethanol has even been proposed as a possible alternative to fossil fuels.

Since 1 mole of ethanol when combusted releases 1367 kJ/mol of energy

x moles of ethanol releases 1418 kJ/mol.

x= 1 × 1418 kJ/mol/ 1367 kJ/mol

x= 1.04 moles of ethanol.

Mass of ethanol = number of moles × molar mass

Molar mass of ethanol = 46.07 g/mol

Mass of ethanol = 1.04 moles × 46.07 g/mol

Mass of ethanol= 47.9 g of ethanol

7 0

2 years ago

A scientist is unsure about the accuracy of her experiment. She has checked her equipment and found it to be in good working ord

polet [3.4K]

Answer:

D

Explanation:

The correct thing to do in this case would be to repeat the experiment.

The scientist would need to repeat the experiment in order to double-check the accuracy. If the accuracy is indeed doubtful, he/she can be able to trace the source of the error by repeating the experiment.

The correct option is D.

8 0

3 years ago

If the molecule could move upward without colliding with other molecules, then how high would it go before coming to rest? Give

tankabanditka [31]

The maximum height at which nitrogen molecule will go before coming to rest is 14 kilometers.

Given:

The nitrogen gas molecule with a temperature of 330 Kelvins is released from Earth's surface to travel upward.

To find:

The maximum height of a nitrogen molecule when released from the Earth's surface before coming to rest.

Solution:

The maximum height attained by nitrogen gas molecule = h
The temperature of nitrogen gas particle = T = 330 K

The average kinetic energy of the gas particles is given by:

$K.E=\frac{3}{2}K_bT\\\\K.E=\frac{3}{2}\times 1.38\times 10^{-23} J/K\times 330 K\\\\K.E=6.381\times 10^{-21} J$

The nitrogen molecule at its maximum height will have zero kinetic energy as all the kinetic energy will get converted into potential energy

The potential energy at height h = $P.E = 6.381\times 10^{-21} J$
Molar mass of nitrogen gas = 28.0134 g/mol
Mass of nitrogen gas molecule = m

$m= \frac{ 28.0134 g/mol}{6.022\times 10^{23} mol^{-1}}=4.652\times 10^{-23} g\\\\1g=0.001kg\\\\m=4.652\times 10^{-23}\times 0.001 kg\\\\=4.652\times 10^{-26} kg$

The acceleration due to gravity = g = 9.8 m/s^2
The maximum height attained by nitrogen gas molecule = h
The potential energy is given by:

$P.E=mgh$

$6.381\times 10^{-21} J=4.652\times 10^{-26} kg\times 9.8 m/s^2\times h\\\\h=\frac{6.381\times 10^{-21} J}{4.652\times 10^{-26} kg\times 9.8 m/s^2}\\\\h=13,996.6 m\\\\1 m = 0.001 km\\\\h=13,996.6 m=h=13,996.6\times 0.001 k m\\\\=13.9966 km \approx 14 km$

The maximum height at which nitrogen molecule will go before coming to rest is 14 kilometers.

Learn more about the average kinetic energy of gas particles here:

brainly.com/question/16615446?referrer=searchResults

brainly.com/question/6329137?referrer=searchResults

6 0

2 years ago

How would you prepare 100 ml of 0.4 M MgSO4 from a stock solution of 2 M MgSO4?

miss Akunina [59]

OK, so to answer this question, you will simply use the molality equation which is as follows:
M1V1 = M2V2
In the givens you have:
M1 = 2M
V1 is the unknown
M2 = 0.4M
V2 = 100 ml

plug in the givens in the above equation:
2 x V1 = 0.4 x 100
therefore:
V1 = 20 ml

Based on this: you should take 20 ml of the 2 M solution and make volume exactly 100 ml in a volumetric flask by diluting in water.

7 0

2 years ago

Alveoli are tiny sacs of air in the lungs. Their average diameter is 4.50 × 10−5 m. Calculate the uncertainty in the velocity of

Arisa [49]

Answer: The uncertainty in the velocity of oxygen molecule is $4.424\times 10^{-5}m/s$

Explanation:

The diameter of the molecule will be equal to the uncertainty in position.

The equation representing Heisenberg's uncertainty principle follows:

$\Delta x.\Delta p=\frac{h}{2\pi}$

where,

$\Delta x$ = uncertainty in position = d = $4.50\times 10^{-5}m$

$\Delta p$ = uncertainty in momentum = $m\Delta v$

m = mass of oxygen molecule = $5.30\times 10^{-26}kg$

h = Planck's constant = $6.627\times 10^{-34}kgm^2/s^2$

Putting values in above equation, we get:

$4.50\times 10^{-5}m\times 5.30\times 10^{-26}kg\times \Delta v=\frac{6.627\times 10^{-34}kgm^2/s}{2\times 3.14}\\\\\Delta v=\frac{6.627\times 10^{-34}kgm^2/s^2}{2\times 3.14\times 4.50\times 10^{-5}m\times 5.30\times 10^{-26}kg}=4.424\times 10^{-5}m/s$

Hence, the uncertainty in the velocity of oxygen molecule is $4.424\times 10^{-5}m/s$

8 0

3 years ago