Answer:
geosphere
Explanation:
geo means rocks this is the area
hope this helps dude
Answer:
Its a physical
theres no chemical properties involved.
Explanation:
Brainliest please
Answer:
81.71%
Explanation:
One mole of propane contains 3 moles of carbon atoms and 8 moles of hydrogen atoms, as seen from the molecular formula of
. In order to calculate the percent of carbon in propane by mass, we need to remember that %w/w (or percent mass) formula states that:

That is, we need to divide the mass of the component of interest by the total mass of the compound and multiply by 100 to obtain the percentage.
For simplicity, let's take 1 mole of propane and find the mass of 1 mole (hence, we'll be finding the molar mass of propane). To do that, we add the 3 molar masses of carbon and 8 molar masses of hydrogen to obtain a total of:

Now that we have the molar mass of propane, we also need to find the total mass of carbon in 1 mole of propane. We know that we have a total of 3 moles of carbon which corresponds to:

Dividing the mass of carbon present by the total mass of the compound will yield the mass percentage as defined by the formula we introduced:

A pendulum. Hold it high and it has potential energy but no kinetic energy because it is still. As it moves down when you drop it, it changes to kinetic energy and moves faster and faster until it's at the bottom where it has no potential energy. However, it still has momentum, so now it is changed to potential energy as it moves up the other side
Answer:
Kc = 3.90
Explanation:
CO reacts with
to form
and
. balanced reaction is:

No. of moles of CO = 0.800 mol
No. of moles of
= 2.40 mol
Volume = 8.00 L
Concentration = 
Concentration of CO = 
Concentration of
= 

Initial 0.100 0.300 0 0
equi. 0.100 -x 0.300 - 3x x x
It is given that,
at equilibrium
= 0.309/8.00 = 0.0386 M
So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M
At equilibrium
= 0.300 - 0.0386 × 3 = 0.184 M
At equilibrium
= 0.0386 M
![Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BH_2O%5D%5BCH_4%5D%7D%7B%5BCO%5D%5BH_2%5D%5E3%7D)
