Answer:
Formula Weight of gas sample = 20.1 g/mole => Neon (Ne)
Explanation:
Use Ideal Gas Law formula to determine formula weight and compare to formula weights of answer choices.
PV = nRT = (mass/fwt)RT => fwt = (mass/Volume)RT = Density x R x T
Density = 0.900 grams/L
R = 0.08206 L·atm/mole·K
T = 0.00°C = 273Kfwt = (0.900g/L)(0.08206L·atm/mole·K )(273K)
= 20.1 g/mol => Neon (Ne)
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Answer:
330.95K
Explanation:
V₁ = 1.2L
T₁ = 25°C = (25 + 273.15)K = 298.15K
P₁ = 1.0 atm
P₂ = 0.74 atm
V₂ = 1.8L
T₂ =?
From combined gas equation,
(P₁ * V₁) / T₁ = (P₂ * V₂) / T₂
Solve for T₂
T₂ = (P₂ * V₂ * T₁) / (P₁ * V₁)
T₂ = (0.74 * 1.8 * 298.15) / (1.0 * 1.2)
T₂ = 397.1358 / 1.2
T₂ = 330.9465K
T₂ = 330.95K or T₂ = (330.95 - 273.15)°C = 57.8°C
Answer:5.309 × 10²⁴ atoms.
Explanation:
Given that
molar mass of NH3 = 17
g/mol
Mass of NH3 = 5g
Therefore, No of moles of NH3 = Mass/ molar mass
= 5g/ 17g/mol
= 0.294 moles.
I mole = 6.02 × 10²³ atoms
Therefore the number of hydrogen atoms in a 0.294 moles of ammonia gives us
0.294× 6.02 × 10²³ × 3 ( since there are 3 hydrogens in Ammonia )
= 5.309 × 10²⁴ atoms.
Unburned hydrocarbon on reacting with oxygen undergoes combustion reaction. However, the activation energy of this reaction is significantly high. When a catalyst like Pd is added to the reaction system, it provides active sites for the reaction to occur. It acts are a heterogeneous catalyst. It is pertinent of note that catalyst is refereed as heterogeneous, when it exist in different phase as compared to reactant and products. In present case, reactants and products are in gas phase, while catalyst is in solid phase. Due to availability of larger surface area at active site of Pd, activation energy of reaction decreases and decrease in activation energy favors higher reaction rates.
