Answer : The concentration of A after 80 min is, 0.100 M
Explanation :
Half-life = 20 min
First we have to calculate the rate constant, we use the formula :
Expression for rate law for first order kinetics is given by:
where,
k = rate constant = 
t = time passed by the sample = 80 min
a = initial amount of the reactant = 1.6 M
a - x = amount left after decay process = ?
Now put all the given values in above equation, we get
Therefore, the concentration of A after 80 min is, 0.100 M
Answer:
B. Hydrogen and oxygen molecules
General Formulas and Concepts:
<u>Chemistry - Reactions</u>
- Reactants are always on the left side of the arrow
- Products are always on the right side of the arrow
Explanation:
<u>Step 1: Define</u>
Reaction RxN: 2H₂ + O₂ → 2H₂O
<u>Step 2: Identify</u>
Reactants: H₂ and O₂
Products: H₂O
Answer:
The enthalpy of the nitrogen-nitrogen bond in N2H4 is 162.6 kJ
Explanation:
For the reaction: N2H4(g)+H2(g)→2NH3(g), the enthalpy change of reaction is
ΔH rxn = 2 ΔHºf NH3 - ΔHºf N2H4
but we also know that the ΔH rxn is calculated by accounting the sum of number of bonds formed and bonds broken as follows:
ΔH rxn = 6H (N-H) + 4 (N-H) + 2H (H-H)
where H is the bond enthalpy .When bonds are broken H is positive, and negative when formed, in the product there are 6 N-H bonds , and in the reactants 4 N-H and 1 H-H bonds).
Consulting an appropiate reference handbook or table the following values are used:
ΔHºf (NH3) = -46 kJ/mol
ΔHºf (N2H4) = 95.94 kJ/mol
(The enthalpy of fomation of hydrogen in its standard state is zero)
H (N-H) = 391 kJ
H (H-H) = 432 kJ
H (N-N) = ?
So plugging our values:
ΔH rxn = 2mol ( -46.0 kJ/mol) - 1mol(95.4 kJ/mol) = -187.40 kJ
-187.40 kJ = 6(-391 kJ) + 4 (391 kJ) + 432 + H(N-N)
-187.40 kJ = -350 kJ + H(N-N)
H(N-N) = 162.6 kJ
Match every A with T and every C with G. Likewise, match T with A and G with C.
