B. normal contact force acts in the opposite direction as of weight, hence A is out. C is out as if normal contact force is lesser than the weight, the object would be sinking into the ground due to net force of the weight. D is out as if normal contact force is greater than the weight, the object would be flying upwards.
Answer:
divergent
Explanation:
I believe it is divergent.
The balanced chemical equation will tell you the molar ratios in which hydrogen and oxygen react to form water:
2H2 + O2 ----> 2H2O
Then the ratio is 2 mol of H2 react with 1 mol of O2 and form 2 mol of H2O:
2 : 1 : 2
Now convert the given data into moles:
16 grams of H2 = 16 g / 2 g/mol = 8 mol
16 grams of O2 = 16 g / 16 g/mol = 1 mol.
Then you have that the 1 mol of oxygen will react with 2 moles of hydrogen and they will form 2 mol of H2O.
And you can calculate the mass of 2 moles of H2O by multiplying by its molar weigth: 18 g/mol.
2 mol H2O * 18 g/mol = 36 grams of water.
Then, the answer is 36 grams of H2O.
The question is incomplete, complete question is;
Carbon monoxide replaces oxygen in oxygenated hemoglobin according to the reaction:
Use the reactions and associated equilibrium constants at body temperature to find the equilibrium constant for the above reaction.
Answer:
The equilibrium constant for the given reaction is 170.
Explanation:
![K_1=\frac{[HbO_2]}{[Hb][O_2]}](https://tex.z-dn.net/?f=K_1%3D%5Cfrac%7B%5BHbO_2%5D%7D%7B%5BHb%5D%5BO_2%5D%7D)
![[Hb]=\frac{[HbO_2]}{[K_1][O_2]}](https://tex.z-dn.net/?f=%5BHb%5D%3D%5Cfrac%7B%5BHbO_2%5D%7D%7B%5BK_1%5D%5BO_2%5D%7D)
..[1]
![K_2=\frac{[HbCO]}{[Hb][CO]}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BHbCO%5D%7D%7B%5BHb%5D%5BCO%5D%7D)
..[2]
![K_c=\frac{[HbCO][O_2]}{[HbO_2][CO]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BHbCO%5D%5BO_2%5D%7D%7B%5BHbO_2%5D%5BCO%5D%7D)
..[3]
Using [1] in [2]:
![K_2=\frac{[HbCO]}{\frac{[HbO_2]}{[K_1][O_2]}\times [CO]}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BHbCO%5D%7D%7B%5Cfrac%7B%5BHbO_2%5D%7D%7B%5BK_1%5D%5BO_2%5D%7D%5Ctimes%20%5BCO%5D%7D)
![K_2=K_1\times \frac{[HbCO][O_2]}{[HbO_2][CO]}](https://tex.z-dn.net/?f=K_2%3DK_1%5Ctimes%20%5Cfrac%7B%5BHbCO%5D%5BO_2%5D%7D%7B%5BHbO_2%5D%5BCO%5D%7D)
( using [3])
The equilibrium constant for the given reaction is 170.
