Answer:
1. The pressure will be 32 atm, twice the initial pressure.
2. The pressure will be 1.83 atm, one third of the initial pressure.
Explanation:
Boyle's law is one of the gas laws that relates the volume and pressure of a certain quantity of gas kept at a constant temperature.
This law says that "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure." This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.
Boyle's law is expressed mathematically as:
Pressure * Volume = constant
or P * V = k
Ahora es posible suponer que tienes un cierto volumen de gas V1 que se encuentra a una presión P1 al comienzo del experimento. Si varias el volumen de gas hasta un nuevo valor V2, entonces la presión cambiará a P2, y se cumplirá:
P1*V1=P2*V2
1. In this case:
- P1= 16 atm
- V1
- P2= ?
- V2= V1÷2=
because the volume is halved.
So:
16 atm*V1= P2* 
Solving:
=P2
16 atm*2= P2
32 atm= P2
<u><em>The pressure will be 32 atm, twice the initial pressure.</em></u>
2. Now
- P1= 5.5 atm
- V1
- P2= ?
- V2= V1*3 because the volume is tripled.
So:
5.5 atm*V1= P2* V1*3
Solving:
=P2
= P2
1.83 atm= P2
<u><em>The pressure will be 1.83 atm, one third of the initial pressure.</em></u>
Answer: Equilibrium constant is 0.70.
Explanation:
Initial moles of
= 0.35 mole
Volume of container = 1 L
Initial concentration of
Initial moles of
= 0.40 mole
Volume of container = 1 L
Initial concentration of
equilibrium concentration of
[/tex]
The given balanced equilibrium reaction is,

Initial conc. 0.35 M 0.40M 0 0
At eqm. conc. (0.35-x) M (0.40-x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[CO_2]\times [H_2O]}{[CO]\times [H_2O]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO_2%5D%5Ctimes%20%5BH_2O%5D%7D%7B%5BCO%5D%5Ctimes%20%5BH_2O%5D%7D)

we are given : (0.35-x)= 0.18
x = 0.17
Now put all the given values in this expression, we get :


Thus the value of the equilibrium constant is 0.70.
You can take two liquids of different densities (how much mass is in a given volume) and pour them into a funnel. An example is oil and water. When the mixture settles, the denser liquid will be at the bottom, and drips through the funnel first. This is a separation that you can just let occur naturally.
The reaction is given as
Fe2O3 (s)+ 3CO(g)--->3CO2(g)+ 2Fe(s)
No.of moles=mass in gram/molar mass
As for Fe mole =156.2g/55.847=2.7969~2.797
The ratio b/w CO and Fe is 3:2
Moles of CO needed= 2.797x3/2=4.1955
Mass of CO needed= 4.195mol x 28.01g/mol= 117.515g
Explanation:
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