5.58 X 
Litres is the volume, in liters, occupied by 0.015 molecules of oxygen at STP.
Explanation:
Data given:
molecules of oxygen = 0.015
number of moles of oxygen =?
temperature at STP = 273 K
Pressure at STP = 1 atm
volume = ?
R (gas constant) = 0.08201 L atm/mole K
to convert molecules to moles,
number of moles = 
number of moles = 2.49 x 
Applying the ideal gas law since the oxygen is at STP,
PV = nRT
rearranging the equation:
V = 
putting the values in the rearranged equation:
V = 
V = 5.58 X Litres.
Answer:
The reaction quotient (Q) before the reaction is 0.32
Explanation:
Being the reaction:
aA + bB ⇔ cC + dD
![Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BC%5D%5E%7Bc%7D%20%2A%5BD%5D%5E%7Bd%7D%20%7D%7B%5BA%5D%5E%7Ba%7D%2A%5BB%5D%5E%7Bb%7D%20%20%7D)
where Q is the so-called reaction quotient and the concentrations expressed in it are not those of the equilibrium but those of the different reagents and products at a certain instant of the reaction.
The concentration will be calculated by:
You know the reaction:
PCl₅ (g) ⇌ PCl₃(g) + Cl₂(g).
So:
![Q=\frac{[PCl_{3} ] *[Cl_{2} ] }{[PCl_{5} ]}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BPCl_%7B3%7D%20%5D%20%2A%5BCl_%7B2%7D%20%5D%20%7D%7B%5BPCl_%7B5%7D%20%5D%7D)
The concentrations are:
- [PCl₃]=
- [Cl₂]=
- [PCl₅]=
Replacing:
Solving:
Q= 0.32
<u><em>The reaction quotient (Q) before the reaction is 0.32</em></u>
Answer:
Final temperature = T₂ = 155.43 °C
Explanation:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Given data:
Mass of coin = 4.50 g
Heat absorbed = 54 cal
Initial temperature = 25 °C
Specific heat of copper = 0.092 cal/g °C
Final temperature = ?
Solution:
Q = m.c. ΔT
ΔT = T₂ -T₁
Q = m.c. T₂ -T₁
54 cal = 4.50 g × 0.092 cal/g °C × T₂ -25 °C
54 cal = 0.414 cal/ °C × T₂ -25 °C
54 cal /0.414 cal/ °C = T₂ -25 °C
130.43 °C = T₂ -25 °C
130.43 °C + 25 °C = T₂
155.43 °C = T₂
The charge of a Rb ion would be +1
I believe the answer to your question is none of the above
