Answer:
Strontium (St)
Explanation:
The atom that will have a larger radius than zinc is strontium, Sr.
Atomic radius is defined as the half of the inter-nuclear distance between two covalently bonded atoms of non-metallic elements or half of the distance between two nuclei in the solid state of metals.
- Across the period atomic radii decreases progressively due to the increase in nuclear charge.
- Down a group atomic radii increase progressively due to the successive shells of electrons being added.
- Since strontium satisfies the criteria, it has the larger atomic radius.
Answer:
1.5 hours or 90 minutes
Explanation:
Velocity = d/t
V = 64 mi / hour or 96 mi / 9 min OR 16 mi / 15 min
D = 96 miles
Velocity * Time = Distance
Time = Distance / Velocity
T = 96 mi / 64 mi / hour
96 / 64 = 1.5 mi per hour
T = 1.5 hours or 90 minutes
<span>Using PV=nRT to find the moles and then convert back.
</span><span>4x=.8944
</span><span>solve for x then use the pressure for lets say CO2 put that into PV=nRT then solve for n then convert over.
</span>
<span>(.2236)(2)/(298*.08206) = .0183*96g/mol = 1.76g
</span>
<span>For C:
[NH3]^2[CO2][H2O] = Kp
x=0.2236
(2*.2236)^2(.2236)*(.2236)
=0.001
</span>
Answer:
3. V = 0.2673 L
4. V = 2.4314 L
5. V = 0.262 L
6. V = 2.224 L
Explanation:
3. assuming ideal gas:
∴ R = 0.082 atm.L/K.mol
∴ V1 = 225 L
∴ T1 = 175 K
∴ P1 = 150 KPa = 1.48038 atm
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(175 K))/((1.48038 atm)(225 L))
⇒ n = 0.043 mol
∴ T2 = 112 K
∴ P2 = P1 = 150 KPa = 1.48038 atm
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(112 K)(0.043 mol))/(1.48038 atm)
⇒ V2 = 0.2673 L
4. gas is heated at a constant pressure
∴ T1 = 180 K
∴ P = 1 atm
∴ V1 = 44.8 L
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(180 K))/((1 atm)(44.8 L))
⇒ n = 0.3295 mol
∴ T2 = 90 K
⇒ V2 = RT2n/P
⇒ V2 = ((0.082 atm.L/K.mol)(90 K)(0.3295 mol))/(1 atm)
⇒ V2 = 2.4314 L
5. V1 = 200 L
∴ P1 = 50 KPa = 0.4935 atm
∴ T1 = 271 K
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(271 K))/((0.4935 atm)(200 L))
⇒ n = 0.2251 mol
∴ P2 = 100 Kpa = 0.9869 atm
∴ T2 = 14 K
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(14 K)(0.2251 mol))/(0.9869 atm)
⇒ V2 = 0.262 L
6.a) ∴ V1 = 24.6 L
∴ P1 = 10 atm
∴ T1 = 25°C = 298 K
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(298 K))/((10 atm)(24.6 L))
⇒ n = 0.0993 mol
∴ T2 = 273 K
∴ P2 = 101.3 KPa = 0.9997 atm
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(273 K)(0.0993 mol))/(0.9997 atm)
⇒ V2 = 2.224 L
