Answer:
Here are a few more examples:
Smoke and fog (Smog)
Dirt and water (Mud)
Sand, water and gravel (Cement)
Water and salt (Sea water)
Potassium nitrate, sulfur, and carbon (Gunpowder)
Oxygen and water (Sea foam)
Petroleum, hydrocarbons, and fuel additives (Gasoline)
Heterogeneous mixtures possess different properties and compositions in various parts i.e. the properties are not uniform throughout the mixture.
Examples of Heterogeneous mixtures – air, oil, and water, etc.
Examples of Homogeneous mixtures – alloys, salt, and water, alcohol in water, etc.
Explanation:
Answer:
The equilibrium concentration of NO is 0.001335 M
Explanation:
Step 1: Data given
The equilibrium constant Kc is 0.0025 at 2127 °C
An equilibrium mixture contains 0.023M N2 and 0.031 M O2,
Step 2: The balanced equation
N2(g) + O2(g) ↔ 2NO(g)
Step 3: Concentration at the equilibrium
[N2] = 0.023 M
[O2] = 0.031 M
Kc = 0.0025 = [NO]² / [N2][O2]
Kc = 0.0025 = [NO]² / (0.023)(0.031)
[NO] = 0.001335 M
The equilibrium concentration of NO is 0.001335 M
Answer:
H2
Explanation:
I would say H2 cause CH4 have less hydrogen gas .
Answer:
0.35 milli moles of ethanol can be theoretically be produced under these conditions.
Explanation:
Moles of glucose = 
milli mole
Moles of ADP = 0.35 milli mole
Moles of Pi = 0.35 milli mole
Moles of ATP = 0.70 milli mole
As we can see that ADP and Pi are in limiting amount which means tat they are limiting reagent. So, the moles of ethanol produced will depend upon the moles of ADP and Pi.
According to reaction, 2 moles of ADP gives 2 moles of glucose.
Then 0.35 milli moles of ADp will give :
of ethanol
0.35 milli moles of ethanol can be theoretically be produced under these conditions.
