B. product
Think about it like this: a company produces products, not coefficients.
Answer:
K
Explanation:
Since the blood moves from the body to the right atrium to the right ventricle
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)
To find this, we will use this formula:
Molar mass of element
------------------------------------ x 100
Molar mass of compound
So, first lets calculate the mass of the compound as a whole. We use the atomic masses on the periodic table to determine this.
Ca: 40.078 g/mol
N2 (there is two nitrogens): 28.014 g/mol
O6 (there are six nitrogens: 3 times 2): 95.994 g/mol
When we add all of those numbers up together, we get 164.086. That is the molar mass for the whole compound. However, we are trying to figure out what percent of the compound oxygen makes up. From the molar mass, we know that 95.994 of the 164.086 is oxygen. Lets plug those numbers into our equation!
95.994
-----------
164.086
When we divide those two numbers, we get .585. When we multiply that by 100, we get 58.5.
So, the percent compostition of oxygen in Ca(NO3)2, or, calcium nitrate, is 58.5%.
