Answer:
I can list four. These are the main ones.
Bolling
Filtration
Distillation.
Chlorination
Explanation:
Answer:
D.Lowering the temperature is the best option.
Explanation:
The value of equilibrium constants aren't changed with change in the pressure or concentrations of reactants and products in equilibrium. The only thing that changes the value of equilibrium constant is a change of temperature.
In the reaction below for example;
A + B <==>C+D
If you have moved the position of the equilibrium to the right (and so increased the amount of C and D), why hasn't the equilibrium constant increased?
Let's assume that the equilibrium constant mustn't change if you decrease the concentration of C - because equilibrium constants are constant at constant temperature. Why does the position of equilibrium move as it does?
If you decrease the concentration or pressure of C, the top of the Kc expression gets smaller. That would change the value of Kc. In order for that not to happen, the concentrations of C and D will have to increase again, and those of A and B must decrease. That happens until a new balance is reached when the value of the equilibrium constant expression reverts to what it was before.
Answer:
Its a or d if im stupid my b
Explanation:
Answer:
The answer to your question is 330 g of CHO
Explanation:
Data
Calories needed = 2200 kcal/day
CHO = 60%
Proteins = 15%
Fats = 25%
Grams of carbohydrates needed = ?
Process
1.- Calculate the number of calories in 60% of 2200 kcal
2200 kcal ---------------- 100%
x --------------- 60%
x = (60 x 2200) / 100
x = 1320 kcal
2.- Calculate the grams of CHO
1 g of CHO ---------------- 4 kcal
x ---------------- 1320 kcal
x = (1320 x 1) / 4
x = 1320/4
x = 330 g of CHO
How many oxygen molecules are in 22.4 liters of oxygen gas
at 273k and 101.3kpa
First solve the number of moles of the oxygen gas by using
the ideal gas equation:
PV = nRT
Where n is the number of moles
n = PV/RT
n = (101 300 Pa) (22.4 L) (1 m3/1000 L ) / ( 8.314 Pa m3 /
mol K) ( 273 K)
n = 1 mol O2
the number of molecules can be solve using avogrados number
6.022x10^23 molecule / mole
molecules of one mole O2 = 6.022x 10^23 molecules
