How many liters will 2.76 mol CO2 occupy? 2.76 mol II
1 answer:
This problem is providing information about the moles of carbon dioxide, 2.76 mol, and asks for the volume this amount takes up, turning out to be 61.8 L according to the Avogadro's law:
<h3>Avogadro's law:</h3><h3 />In chemistry, gas laws are used to relate the behavior of gases by virtue of the their pressure, volume, temperature and moles; thus several gas laws exist for us to do so, however, we here focus on the Avogadro's law which relates the volume and moles when both temperature and pressure are held constant.
In such a way, since no information on the constant variables is given, we assume the mentioned carbon dioxide is at STP, (0 °C and 1 atm), which means that we can use the following equivalence statement derived from the ideal gas law (PV=nRT):
22.4 L = 1 mol
Hence, we calculate the required volume:
Learn more about ideal gases: brainly.com/question/11676583
You might be interested in
Answer:
the Molar heat of Combustion of diphenylacetylene =
Explanation:
Given that:
mass of diphenylacetylene = 0.5297 g
Molar Mass of diphenylacetylene = 178.21 g/mol
Then number of moles of diphenylacetylene =
=
= 0.002972 mol
By applying the law of calorimeter;
Heat liberated by 0.002972 mole of diphenylacetylene = Heat absorbed by
+ Heat absorbed by the calorimeter
Heat liberated by 0.002972 mole of diphenylacetylene = msΔT + cΔT
= 1369 g × 4.184 J g⁻¹°C⁻¹ × (26.05 - 22.95)°C + 916.9 J/°C (26.05 - 22.95)°C
= 17756.48 J + 2842.39 J
= 20598.87 J
Heat liberated by 0.002972 mole of diphenylacetylene = 20598.87 J
Heat liberated by 1 mole of diphenylacetylene will be =
= 6930979.139 J/mol
= 6930.98 kJ/mol
Since heat is liberated ; Then, the Molar heat of Combustion of diphenylacetylene =
Answer:
6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.
Explanation:
Using integrated rate law for first order kinetics as:
Where,
is the concentration at time t
is the initial concentration
Given:
Concentration is decreased to 1.56 % which means that 0.0156 of is decomposed. So,
= 0.0156
Thus,
kt = 4.1604
The expression for the half life is:-
Half life = 15.0 hours
Where, k is rate constant
So,
<u>6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.</u>