Answer:
3NaOH+FeBr3>3NaBr+
Fe(OH)3
Explanation:
After writing the equation it has to be balanced
Answer:
The amount of energy released from the combustion of 2 moles of methae is 1,605.08 kJ/mol
Explanation:
The chemical reaction of the combustion of methane is given as follows;
CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g)
Hence, 1 mole of methane combines with 2 moles of oxygen gas to form 1 mole of carbon dioxide and 2 moles of water vapor
Where:
CH₄ (g): Hf = -74.6 kJ/mol
CO₂ (g): Hf = -393.5 kJ/mol
H₂O (g): Hf = -241.82 kJ/mol
Therefore, the combustion of 1 mole of methane releases;
-393.5 kJ/mol × 1 + 241.82 kJ/mol × 2 + 74.6 kJ/mol = -802.54 kJ/mol
Hence the combustion of 2 moles of methae will rellease;
2 × -802.54 kJ/mol or 1,605.08 kJ/mol.
The net force is the sum of all of the forces that cancelled out to zero. Hence, option D is correct.
<h3>What is force?</h3>
The pushes or pulls on an object with mass causes it to change its velocity.
The net force is the vector sum of all forces acting on an object. Force is a vector quantity.
Two forces of equal magnitude when acting on an object in opposite directions will cancel out each other.
As a result net force is equal to 0. The word net here shows the total force acting on the object.
Hence, option D is correct.
Learn more about the force here:
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Answer:
Eutrophication is the enrichment of a body of water with excessive nutrients (nitrogen and phosphorus), which causes algal growth and subsequent decline of dissolved oxygen after decomposition.
2.0 L
The key to any dilution calculation is the dilution factor
The dilution factor essentially tells you how concentrated the stock solution was compared with the diluted solution.
In your case, the dilution must take you from a concentrated hydrochloric acid solution of 18.5 M to a diluted solution of 1.5 M, so the dilution factor must be equal to
DF=18.5M1.5M=12.333
So, in order to decrease the concentration of the stock solution by a factor of 12.333, you must increase its volume by a factor of 12.333by adding water.
The volume of the stock solution needed for this dilution will be
DF=VdilutedVstock⇒Vstock=VdilutedDF
Plug in your values to find
Vstock=25.0 L12.333=2.0 L−−−−−
The answer is rounded to two sig figs, the number of significant figures you have for the concentration od the diluted solution.
So, to make 25.0 L of 1.5 M hydrochloric acid solution, take 2.0 L of 18.5 M hydrochloric acid solution and dilute it to a final volume of 25.0 L.
IMPORTANT NOTE! Do not forget that you must always add concentrated acid to water and not the other way around!
In this case, you're working with very concentrated hydrochloric acid, so it would be best to keep the stock solution and the water needed for the dilution in an ice bath before the dilution.
Also, it would be best to perform the dilution in several steps using smaller doses of stock solution. Don't forget to stir as you're adding the acid!
So, to dilute your solution, take several steps to add the concentrated acid solution to enough water to ensure that the final is as close to 25.0 L as possible. If you're still a couple of milliliters short of the target volume, finish the dilution by adding water.
Always remember
Water to concentrated acid →.NO!
Concentrated acid to water →.YES!
