Answer:
The molar mass of the unknown gas is 100.4 g/mol
Explanation:
Step 1: Data given
Molar mass of argon = 39.95 g/mol
After filling with argon the flask gained 3.221 grams
After filling with an unknown gas, the flask gained 8.107 grams
Step 2: Calculate the molar mass of the unknown gas
The gas with the higher molar mass will have the higher density.
Ar - 3.224 g; molar mass = 39.95 g/mol
X = 8.102 g; molar mass = ??
Molar mass of the unknown gas = 8.102g X *(39.95 g/mol / 3.224 g) = 100.4 g/mol
The molar mass of the unknown gas is 100.4 g/mol
The answer would be A) cells!!
Answer:
The upper and lower limits for the room-temperature thermal conductivity of a magnesium oxide material having a volume fraction of 0.10 of pores that are filled with still air are
Ku = 38.252 W/mK
K lower = 0.199 W/mK
Explanation:
As we know
Ku = Vp * Kair + Vmagnesium * K metal
Ku = 0.10 *0.02 + (1-0.25) * 51
Ku = 38.252 W/mK
The lower limit
K lower = Kmetal* Kair/( Vp * Kmetal + Vmetal * K air)
K lower = (0.02*51)/(0.10*51 + 0.90 * 0.02)
K lower = 0.199 W/mK
<u>Given:</u>
The initial energy of the electron Einitial = 16.32 * 10⁻¹⁹ J
The energy released i.e the change in energy ΔE = 5.4 * 10⁻¹⁹ J
<u>To determine:</u>
The final energy state Efinal of the electron
<u>Explanation:</u>
Since energy is being released, this suggests that Efinal < Einitial
i.e. ΔE = Einitial - Efinal
Efinal = Einitial - ΔE = (16.32 - 5.4)*10⁻¹⁹ = 10.92 * 10⁻¹⁹ J
Ans: A)
The electron moved down to an energy level and has an energy of 10.92 * 10⁻¹⁹ J
