Unless you're able to provide a diagram representing the problem, there's not much I can do so solve this problem.
The correct answer for the question that is being presented above is this one: "<span>0.3."
Here it is how to solve.
M</span><span>olecular mass of Ar = 40
</span><span>Molecular mass of Ne = 20
</span><span>Number of moles of Ar = 9.59/40 = 0.239
</span><span>Number of moles of Ne = 11.12/20= 0.556
</span><span>Mole fraction of argon = 0.239/ ( 0.239 + 0.556) = 0.3</span><span>
</span>
The rate constant is mathematically given as
K2=2.67sec^{-1}
<h3>What is the Arrhenius equation?</h3>
The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:
Therefore
KT1= 0.0110^{-1}
T1= 21+273.15
T1= 294.15K
T2= 200
T2=200+273.15
T2= 473.15K
Ea= 35.5 Kj/Mol
Hence, in j/mol R Ea is
Ea=35.5*1000 j/mol R
K2/0.0110 =e^(5.492)
K2/0.0110 =242.74
K2= 242.74*0.0110
K2=2.67sec^{-1}
In conclusion, rate constant
K2=2.67sec^{-1}
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Answer:
Explanation:
Oxygen is one of the most abundant elements on this planet. Our atmosphere is 21% free elemental oxygen. Oxygen is also extensively combined in compounds in the earths crust, such as water (89%) and in mineral oxides. Even the human body is 65% oxygen by mass.
Free elemental oxygen occurs naturally as a gas in the form of diatomic molecules, O2 (g). Oxygen exhibits many unique physical and chemical properties. For example, oxygen is a colorless and odorless gas, with a density greater than that of air, and a very low solubility in water. In fact, the latter two properties greatly facilitate the collection of oxygen in this lab. Among the unique chemical properties of oxygen are its ability to support respiration in plants and animals, and its ability to support combustion.
In this lab, oxygen will be generated as a product of the decomposition of hydrogen peroxide. A catalyst is used to speed up the rate of the decomposition reaction, which would otherwise be too slow to use as a source of oxygen. The catalyst does not get consumed by the reaction, and can be collected for re-use once the reaction is complete. The particular catalyst used in this lab is manganese(IV) oxide.
Answer:
40.4 kJ
Explanation:
Step 1: Given data
- Heat of sublimation of CO₂ (ΔH°sub): 32.3 kJ/mol
Step 2: Calculate the moles corresponding to 55.0 g of CO₂
The molar mass of CO₂ is 44.01 g/mol.
n = 55.0 g × 1 mol/44.01 g = 1.25 mol
Step 3: Calculate the heat (Q) required to sublimate 1.25 moles of CO₂
We will use the following expression.
Q = n × ΔH°sub
Q = 1.25 mol × 32.3 kJ/mol = 40.4 kJ
