Answer:
Option (1) Br– is the catalyst, and the reaction follows a faster pathway with Br– than without
Explanation:
Let us consider the equation below:
Step 1:
H2O2(aq) + Br–(aq) → H2O(l) + BrO–(aq)
Step 2:
BrO–(aq) + H2O2(aq) → H2O(l) + O2(g) + Br–(aq)
From the above equation, we can see that Br– is unchanged.
This implies that Br– is the catalyst as catalyst does not take part in a chemical reaction but they create an alternate pathway to lower the activation energy in order for the reaction to proceed at a much faster rate to arrive at the products.
Answer:30 L
Explanation:
Initial Volume
=
V
1
=
60
l
i
t
e
r
Initial Temperature
=
T
1
=
546
K
Final Temperature
=
T
2
=
273
K
Final Vloume
=
V
2
=
?
?
Sol:-
Since the pressure is constant and the question is asking about temperature and volume, i.e,
V
1
T
1
=
V
2
T
2
⇒
V
2
=
V
1
⋅
T
2
T
1
=
60
⋅
273
546
=
60
2
=
30
l
i
t
e
r
⇒
V
2
=
30
l
i
t
e
r
Hence the new volume of the gas is
30
l
i
t
e
r
Let initially there are 10 molecules of O2 and 3 molecules of C3H8 present
The reaction will be
C3H8(g) + 5O2(g) ----> 3CO2(g) + 4H2O
so here oxygen molecules are limiting as for 3 molecules of C3H8 we need 15 molecules of O2
now the given 10 molecules of O2 will react with only 2 molecules of C3H8 and they will form six molecules of CO2 and 8 molecules of H2O
Hence answer is
molecules of CO2 formed = 6
Molecules of H2O formed = 8
molecules of C3H8 left = 1
molecules of O2 left = 0
Answer:
d. The gold(III) ion is most easily reduced.
Explanation:
The standard reduction potentials are
Au³⁺ + 3e⁻ ⟶ Au; 1.50 V
Hg²⁺ + 2e⁻ ⟶ Hg; 0.85 V
Zn²⁺ + 2e⁻ ⟶ Zn; -0.76 V
Na⁺ + e⁻ ⟶ Na; -2.71 V
A <em>more positive voltage</em> means that there is a <em>stronger driving force</em> for the reaction.
Thus, Au³⁺ is the best acceptor of electrons.
Reduction Is Gain of electrons and, Au³⁺ is gaining electrons, so
Au³⁺ is most easily reduced.
Answer:
Oxygen in hydrogen peroxide oxidizes from -1 to 0.
Explanation:
Oxidation is the loss of electrons. The specie which is oxidized has has elevation in its oxidation state as compared in the reactant and the products.
The given reaction is shown below as:
Manganese in 
has oxidation state of +7
Manganese in 
has an oxidation state of +2
It reduces from +7 to +2
Oxygen in hydrogen peroxide has an oxidation state of -1.
Oxygen in molecular oxygen has an oxidation of 0.
Thus, oxygen in hydrogen peroxide oxidizes from -1 to 0.
