The volume of a sample of nitrogen is 6.00 liters at 35oC and 0.75 atm. What volume will it occupy at STP?
1 answer:
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Answer: The statement, average kinetic energy of the gas particles is greater in container A because its particles move faster is correct.
Explanation:
Kinetic energy is the energy obtained due to the motion of an object or substance.
where,
T = temperature
This means that kinetic energy is directly proportional to temperature.
So, when heat is provided to container A then its molecules will start to move rapidly from one place to another which will cause more collisions between the atoms.
Hence, average kinetic energy will be more in container A.
Whereas container B is placed at room temperature which is low than that in container A. So, molecules in container B will move at almost same speed and therefore, specific collisions will be there. So, average kinetic energy in container B will be less than that in container A.
Thus, we can conclude that the statement, average kinetic energy of the gas particles is greater in container A because its particles move faster is correct.
Answer:
The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).
Explanation:
The reaction between an acid and a base is called neutralization, forming a salt and water.
Salt is an ionic compound made up of an anion (positively charged ion) from the base and a cation (negatively charged ion) from the acid.
When an acid is neutralized, the amount of base added must equal the amount of acid initially present. This base quantity is said to be the equivalent quantity. In other words, at the equivalence point the stoichiometry of the reaction is exactly fulfilled (there are no limiting or excess reagents), therefore the numbers of moles of both will be in stoichiometric relationship. So:
V acid *M acid = V base *M base
where V represents the volume of solution and M the molar concentration of said solution.
In this case:
- V acid= 13.7 mL= 0.0137 L (being 1,000 mL= 1 L)
- M acid= 0.129 M
- V base= ?
- M base= 0.135 M
Replacing:
0.0137 L* 0.129 M= V base* 0.135 M
Solving:
V base=0.0131 L = 13.1 mL
<u><em> The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).</em></u>
Answer:
you can solve the rest of the equation. I only reduced it to that much to show you how to derive it
