B. This is because the Hydrogen and Oxygen need balanced out.
Current-
C-1 | C-1
H-4 | H-2
O-2 | O-3
Adding a coefficient of 2 before oxygen in the reactants and H2O in the products would balance this equation
<span>CH4 + 2O2 → CO2 + 2H2O</span>
C-1 | C-1
H-4 | H-4
O-4 | O-4
Answer:
1. Equivalence point
2. Direct titration
3. Primary standard
4. Titrand
5. Back titration
6. Standard solution
7. Titrant
8. Indirect titration
9. End point
10. Indicator
Explanation:
1. The equivalence point is the tiration point at which the quantity or moles of the added titrant is sufficient or equal to the quantity or moles of the analyte for the neutralization of the solution of the analyte.
2. Direct titration is a method of quantitatively determining the contents of a substance
3. A primary standard is an easily weigh-able representative of the mount of moles contained in a substance
4. A titrand is the substance of unknown concentration which is to be determined
5. The titration method that uses a given amount of an excess reagent to determine the concentration of an analyte is known as back titration
6. A standard solution is a solution of accurately known concentration
7. A titrant is a solution that has a known concentration and which is titrated unto another solution to determine the concentration of the second solution
8. Indirect titration is the process of performing a titration in athe reverse order
9. The end point is the point at which the indicator indicates that the equivalent quantities of the reagents required for a complete reaction has been added
10 An indicator is a compound used to visually determine the pH of a solution.
In descending order from top:
E
F
D
A
C
B
All you really need to do is remember the symbols of each, and you’ve got it.
Answer:
K = 3.37
Explanation:
2 NH₃(g) → N₂(g) + 3H₂(g)
Initially we have 4 mol of ammonia, and in equilibrium we have 2 moles, so we have to think, that 2 moles have been reacted (4-2).
2 NH₃(g) → N₂(g) + 3H₂(g)
Initally 4moles - -
React 2moles 2m + 3m
Eq 2 moles 2m 3m
We had produced 2 moles of nitrogen and 3 mol of H₂ (ratio is 2:3)
The expression for K is: ( [H₂]³ . [N₂] ) / [NH₃]²
We have to divide the concentration /2L, cause we need MOLARITY to calculate K (mol/L)
K = ( (2m/2L) . (3m/2L)³ ) / (2m/2L)²
K = 27/8 / 1 → 3.37
