Answer:
eg=linear, mg=linear
Explanation:
First of all, it must be stated that most triatomic molecules are either linear or bent. This depends on the electron geometry of the molecule and the number of bonding groups, multiple bonds and lone pairs present.
CO2 contains four regions of electron density and two bonding groups. For a specie containing two bonding groups, a linear molecular geometry is expected with an angle of 180°.
For a specie having two bonding groups and no lone pairs with multiple bonds, the expected electron geometry is also linear.
Answer:
The concentrations change to maintain the original value of K.
Explanation:
The concentration changes to maintain the original value of K when the concentration of the system changes.
This is the effect concentration has on a reaction at equilibrium.
Le Chatelier's principle proposes that "if any conditions of a system in equilibrium is changed, the system will adjust itself in order to annul the effect of the change".
- The equilibrium constant K is temperature dependent.
- An increase in concentration of a specie favors the direction that uses it up and vice versa.
- This does not change the value of the equilibrium constant.
If a liquid does not have a constant boiling point but it looks the same throughout, then the liquid is probably a homogeneous mixture. A liquid that does not have a constant boiling point would mean that it is not a pure liquid, it would be a mixture of different substances. As stated, the liquid is observed to be the same all throughout then it most likely is homogeneous. Homogeneous mixture are mixture which has the same composition of the components at any point of the mixture. So, it would have uniform properties as well. Examples of this type of mixture would be soda, coffee, tea and juices.<span />
Answer:
C)
the potential energy is increasing through steps A & B, then decreases at C.
Explanation:
Answer:
5.3 atm
Explanation:
We can solve this problem using the equation of state for an ideal gas, which states that:
where
p is the pressure of the gas
V is its volume
n is the number of moles
R is the gas constant
T is the absolute temperature of the gas
For a gas undergoing a transformation, n and R remain constant, so the equation can be written as
For the gas in this problem we have:
is the initial temperature of the gas
is the final temperature
is the initial pressure
, as the volume is decreased by 40.0%
Solving for , we find the final pressure of the gas: