Answer:
A) 6.48 g of OF₂ at the anode.
Explanation:
The gas OF₂ can be obtained through the oxidation of F⁻ (inverse reaction of the reduction presented). The standard potential of the oxidation is the opposite of the standard potential of the reduction.
H₂O(l) + 2 F⁻(aq) → OF₂(g) + 2 H⁺(aq) + 4 e⁻ E° = -2.15 V
Oxidation takes place in the anode.
We can establish the following relations:
- 1 Faraday is the charge corresponding to 1 mole of e⁻.
- 1 mole of OF₂ is produced when 4 moles of e⁻ circulate.
- The molar mass of OF₂ is 54.0 g/mol.
The mass of OF₂ produced when 0.480 F pass through an aqueous KF solution is:
Answer:
distance = 7 miles
displacement = 5 miles
Explanation:
Distance is a scalar quantity as it takes account of magnitude traveled but not the direction traveled from starting point.
The distance traveled is the sum total of distances moved
distance = 4 + 3 = 7 miles
Displacement however, is a vector and measure the shortest possible distance traveled in a given direction from the starting point.
The path of Elvis' walking forms a right-angle triangle with the hypotenuse being the displacement and the other two sides being the distance traveled west and south.
Using Pythagoras' theorem; c² = a² + b²
where c = hypotenuse and a and b are the other two sides
c² = 4² + 3²
c² = 16 + 9 = 25
√c² = √25
c = 5
Therefore, displacement = 5 miles
Answer:
1. 
, remain the same.
2. 
,increase.
3. 
, decrease.
Explanation:
Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.
This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.
Increase the pressure
If the pressure of the container is increased, the volume will decrease according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease in pressure is taking place.
1.
On increasing the pressure ,the equilibrium will shift in direction where decrease in pressure is taking place,so reaction will shift in no direction because number of gaseous moles are same on both sides. Hence no increase in number of moles of products.
2.
On increasing the pressure ,the equilibrium will shift in direction where decrease in pressure is taking place,so reaction will shift in forward direction because number of gaseous moles are less on product sides. Hence ,increase in number of moles of products.
3.
On increasing the pressure ,the equilibrium will shift in direction where decrease in pressure is taking place,so reaction will shift in backward direction because number of gaseous moles are less on reactant side both sides. Hence, decrease in number of moles of products.
