Answer:
a.) 22.4 L Ne.
Explanation:
It is known that every 1.0 mol of any gas occupies 22.4 L.
For the options:
<em>It represents </em><em>1.0 mol of Ne.</em>
<em />
using cross multiplication:
1.0 mol occupies → 22.4 L.
??? mol occupies → 20 L.
The no. of moles of (20 L) Ar = (1.0 mol)(20 L)/(22.4 L) = 0.8929 mol.
using cross multiplication:
1.0 mol occupies → 22.4 L.
??? mol occupies → 2.24 L.
<em>The no. of moles of (2.24 L) Xe </em>= (1.0 mol)(2.24 L)/(22.4 L) = <em>0.1 mol.</em>
- So, the gas that has the largest number of moles at STP is: a.) 22.4 L Ne.
<span>In order to covert a unit, you must know certain number of conversions. In here, the conversion is in unit of length. One meter is equal to 1000 meter. So if it is in cubic form, then the answer of one meter cube is also equal to 1000 cube. Then,
32 m</span>³ (1000 mm/1m)³
<span>or
</span>32 m³ (1000³ mm³/1 m³)
= 3.2 x 10¹⁰ mm³
1. Triassic
2. eon > era > period
3. The scientists study fossils and rock layers to find major changes.
Kinetic Energy (Translational, in all directions) of a gas as per Kinetic theory of gases at abs. Temperature T:
per molecule: KE = 3/2 * k_B T
per mole: KE = 3/2 * R * T
k_B = Boltzmann's constant. R= Universal gas constant.
Kinetic Energy of a gas (as per mechanics)
KE = 1/2 m (Vrms)²
or KE = 1/2 M (Vrms)²
m = mass of a molecule, M = Molar mass of the gas.
Hence, (Vrms)² = 3 k_B T /m = 3 R T /M
R = 8.3 J/°°K/mole
M = 44 gm/mole for CO2 = 0.044 kg/mole
(Vrms)² = 3 * 8.314 J/°K/mole * (273+27)° / (0.044 kg/mole)
Vrms = 412.38 m/sec
Final temperature if the average specific heat capacity of the products is 2.5 J/g·°C is 1302.57°C
Given reaction:
C₂H₄O(g) → CH₄(g) + CO(g)
Now,
Given data for this reaction is :
ΔH°f =-77.4 kJ/mol
ΔH°vaporization =569.4 J/g
= 569.4J/g × 44.0g/mol
ΔH°vaporization= 25.082 kJ/mol
Now,
ΔH°vaporization = H(vapor) - H(liquid)
25.082 = H(vapor) + 77.4 kJ/mol
H(vapor) = -52.318 kJ/mol
Again,
ΔH°rxn = ( H( CO(g))) + (H(CH₄(g))) - (H( C₂H₄O(g)))
= -110.5 - 74.87 + 52.318
ΔH°rxn = -133.053 kJ/mol
Now,
Given
T₁= 93°C = 366K
We know that,
Q= mcΔT
133.053 × 10³ = 44 × 2.5 × (T₂ - 93)
T₂ = 1302.57°C
Thus, from the above conclusion we can say that , Final temperature if the average specific heat capacity of the products is 2.5 J/g·°C is 1302.57°C .
Learn more about Ethylene oxide here: brainly.com/question/14638014
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