Answer:
The average rate of the reaction in terms of disappearance of A is 0.0004 M/s.
Explanation:
Average rate of the reaction is defined as ratio of change in concentration of reactant with respect to given interval of time.
![R_{avg}=-\frac{[A]_2-[A]_1}{t_2-t_1}](https://tex.z-dn.net/?f=R_%7Bavg%7D%3D-%5Cfrac%7B%5BA%5D_2-%5BA%5D_1%7D%7Bt_2-t_1%7D)
Where :
= initial concentration of reactant at 
.
= Final concentration of reactant at 
.
2A+3B → 3C+2D
![R_{avg}=-\frac{1}{2}\frac{[A]_2-[A]_1}{t_2-t_1}](https://tex.z-dn.net/?f=R_%7Bavg%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B%5BA%5D_2-%5BA%5D_1%7D%7Bt_2-t_1%7D)
The concentration of A at (
) = 
The concentration of A at (
) = 
The average rate of reaction in terms of the disappearance of reactant A in an interval of 0 seconds to 20 seconds is :
The average rate of the reaction in terms of disappearance of A is 0.0004 M/s.
Answer:
MnCO3 + 2H2O ⇄ MnO2 + HCO3- + 2e- +3H+
Explanation:
<u>The</u> unbalanced equation
MnCO3 ⇄ MnO2 + HCO3-
In MnCO3, the oxidation number of Mn is +2
In Mno2, the oxidation number of Mn is +4
The change from +2 to +4 requires an addition of 2 electrons (to the right side).
MnCO3 ⇄ MnO2 + HCO3- + 2e-
The total charge now is -3 on the right side. To balance this we add 3 hydrogen atoms on the right side.
MnCO3 ⇄ MnO2 + HCO3- + 2e- +3H+
On the right side we have 4 hydrogen atoms in total. On the left side we have 0 hydrogen atoms. So to balance, we have to add 2H2O on the left side
MnCO3 + 2H2O ⇄ MnO2 + HCO3- + 2e- +3H+
Now the reaction is balanced.
Answer:
V O2 = 1.623 L
Explanation:
- 1 mol ≡ 6.022 E23 molecules
∴ molecules O2 = 4.00 E22 molecules
⇒ moles O2 = (4.00 E22 molecules O2)×(mol O2/6.022 E23 molecules)
⇒ moles O2 = 0.0664 moles
at STP:
∴ T = 25°C ≅ 298 K
∴ P = 1 atm
assuming ideal gas:
∴ V = RTn/P
⇒ V O2 = ((0.082 atm.L/K.mol)(298 K)(0.0664 mol))/( 1 atm)
⇒ V O2 = 1.623 L
Answer
Rapid Combustion
Explanation:
An explosion is a combustion and a quick reaction, That makes it rapid
Answer:
6.4 × 10^-10 M
Explanation:
The molar solubility of the ions in a compound can be calculated from the Ksp (solubility constant).
CaF2 will dissociate as follows:
CaF2 ⇌Ca2+ + 2F-
1 mole of Calcium ion (x)
2 moles of fluorine ion (2x)
NaF will also dissociate as follows:
NaF ⇌ Na+ + F-
Where Na+ = 0.25M
F- = 0.25M
The total concentration of fluoride ion in the solution is (2x + 0.25M), however, due to common ion effect i.e. 2x<0.25, 2x can be neglected. This means that concentration of fluoride ion will be 0.25M
Ksp = {Ca2+}{F-}^2
Ksp = {x}{0.25}^2
4.0 × 10^-11 = 0.25^2 × x
4.0 × 10^-11 = 0.0625x
x = 4.0 × 10^-11 ÷ 6.25 × 10^-2
x = 4/6.25 × 10^ (-11+2)
x = 0.64 × 10^-9
x = 6.4 × 10^-10
Therefore, the molar solubility of CaF2 in NaF solution is 6.4 × 10^-10M
