Answer: The molar mass of H2S is greater than the molar mass of NH3, making the velocity and effusion rate of NH3 particles faster.
Effusion rate is inversely proportional to molar mass.
NH3 will have a higher average molecule velocity, so it will diffuse faster and will reach the other side of the room more quickly.
Explanation: change up your response a bit
Answer:
Kb = 0.428 m/°C
Explanation:
To solve this problem we need to use the <em>boiling-point elevation formula</em>:
- <em>Tsolution</em> - <em>Tpure solvent</em> = Kb * m
Where <em>Tsolution</em> and <em>Tpure solvent</em> are the boiling point of the CS₂ solution (47.52 °C) and of pure CS₂ (46.3 °C), respectively. Kb is the constant asked by the problem, and m is the molality of the solution.
So in order to use that equation and solve for Kb, first we <em>calculate the molality of the solution</em>.
molality = mol solute / kg solvent
- Density of CS₂ = 1.26 g/cm³
- Mass of 410.0 mL of CS₂ ⇒ 410 cm³ * 1.26 g/cm³ = 516.6 g = 0.5166 kg
molality = 0.270 mol / 0.5166 kg = 0.5226 m
Now we <u>solve for Kb</u>:
<em>Tsolution</em> - <em>Tpure solvent</em> = Kb * m
- 47.52 °C - 46.3 °C = Kb * 0.5226 m
The ideal gas equation is;
PV = nRT; therefore making P the subject we get;
P = nRT/V
The total number of moles is 0.125 + 0.125 = 0.250 moles
Temperature in kelvin = 273.15 + 18 = 291.15 K
PV = nRT
P = (0.250 × 0.0821 )× 291.15 K ÷ (7.50 L) = 0.796 atm
Thus, the pressure in the container will be 0.796 atm
Answer:
Explanation:
q= mc theta
where,
Q = heat gained
m = mass of the substance = 670g
c = heat capacity of water= 4.1 J/g°C
theta =Change in temperature=(
66-25.7)
Now put all the given values in the above formula, we get the amount of heat needed.
q= mctheta
q=670*4.1*(66-25.7)
=670*4.1*40.3
=110704.1
Answer: CuI₂ + Br₂
Explanation:
1) The activity series F > Cl > Br > I means that F is the most active and I is the least active of those four elements (the halogens, group 17 in the periodic table).
The activity is a measure of how eager is an element to react compared to other elements in the series in a single replacement reaction.
2) Choice 1: CuI₂ + Br₂
Since the activity of Br is higher than that of I, Br will react with CuI₂, displacing I, which will be left alone, as per this chemical equation:
CuI₂ + Br₂ → CuBr₂ + I₂
Being I less active than Br, it cannot displace Br in CuBr₂.
3) Choice 2: Cl₂ + AlF₃
Being Cl less active than F, the former will not displace the latter, and the reaction will not proceed.
4) Choice 3: Br₂ + NaCl
Again, being Br less active than Cl, the former will not displace the latter, and the reaction will not proceed.
5) Choice 4: CuF₂ + I₂
Once more, being I less active than F, the former will not displace the latter, and the reaction will not proceed.
