Answer:
The O is being oxidized, but at the same time, is being reducted.
Explanation:
H₂O₂(l) + ClO₂(aq) → ClO₂(aq) + O₂(g)
In this reaction, we have 4 compounds:
Hydrogen peroxide
Chlorine dioxide (twice)
Oxygen
In both dioxide, the Cl acts with +4 in oxidation state; the oxygen acts with -2.
Oxgen in ground state has 0, as oxidation number.
In peroxide, the H acts with +1 but the oxygen acts with -1.
Peroxide is making the oxidation number from the O in the ClO₂, to decrease (reduction) and to increase in the O, at the ground state.
Hydrogen peroxide is a good reducing and oxidizing agent at the same time.
Answer:
P₂ = 140 KPa
Explanation:
Given data:
Initial volume = 8.0 L
Final volume = 4.0 L
Initial pressure = 70 KPa
Final pressure = ?
Solution:
According to Boyle's law
P₁V₁ = P₂V₂
P₂ = P₁V₁ / V₂
P₂ = 70 KPa ×8.0 L/4.0 L
P₂ = 560 KPa .L / 4.0 L
P₂ = 140 KPa
Answer:
Option A. It has stayed the same.
Explanation:
To answer the question given above, we assumed:
Initial volume (V₁) = V
Initial temperature (T₁) = T
Initial pressure (P₁) = P
From the question given above, the following data were:
Final volume (V₂) = 2V
Final temperature (T₂) = 2T
Final pressure (P₂) =?
The final pressure of the gas can be obtained as follow:
P₁V₁/T₁ = P₂V₂/T₂
PV/T = P₂ × 2V / 2T
Cross multiply
P₂ × 2V × T = PV × 2T
Divide both side by 2V × T
P₂ = PV × 2T / 2V × T
P₂ = P
Thus, the final pressure is the same as the initial pressure.
Option A gives the correct answer to the question.
The moles which were measured out is calculated using the following formula
moles = mass/molar mass
molar mass of CuBr2.4H20 = 63.5 Cu + ( 2 x79.9) br + ( 18 x4_) h20 = 295.3 g/mol
moles is therefore= 5.2 g/ 295.3 g/mol= 0.0176 moles
I believe that would be Cadmium Permaganate
Hope that helped :)
