Answer:
I don't really get the options but it favoures the reactant side.
Explanation:
Increasing pressure favours the side with fewer moles of gas while decreasing pressure favours the side with the more moles of gas. E.g
If there is 0 moles of gas particles in the reactant side and 1 mole of gas particle in the product side, increasing pressure favours the reactants while decreasing pressure favours the product side.
With the explanations I have made, I hope the question is now clear to you.
X ml - <span>25% alcohol mixture
y ml - </span><span>90% alcohol mixture
x+y = 455
0.25x ml alcohol in </span>x ml of 25% alcohol mixture
0.9y ml alcohol in y ml of 90% alcohol mixture
0.75*455= 341.25 ml alcohol in 455 ml of 75% alcohol mixture
0.25x+0.9y=341.25
System of equations:
x+y = 455 /*(-0.25) ------> -0.25x-0.25y = -0.25*455
0.25x+0.9y=341.25
-0.25x-0.25y=-113.75
0.25x+0.9y=341.25 Add both equations
0.25x+0.9y-0.25x-0.25y=341.25-113.75
0.65y =227.5
y=227.5/0.65 = 350 ml of 90% alcohol mixture
x+y=455
x+355=455
x=100 ml of 25% alcohol mixture
the answer is A. SnCl2 . 2H2O
Answer:
About 25 kPa
Explanation:
Pressure decreases with height above sea level.
The calculation is rather complicated, so I will refer to the figure below.
It shows that the pressure at 10 km is about 25 kPa.
Answer:
The oxidizing agent is the MnO₄⁻
Explanation:
This is the redox reaction:
10 I⁻ (aq) + 2 MnO₄⁻ (aq) + 16 H⁺ (aq) → 5 I₂ (s) + 2 Mn²⁺ (aq) + 8 H2O (l)
Let's determine the oxidation and the reduction.
I⁻ acts with -1 in oxidation state and changes to 0, at I₂.
All elements in ground state has 0 as oxidation state.
As the oxidation state has increased, this is the oxidation, so the iodide is the reducing agent.
In the permanganate (MnO₄⁻), Mn acts with +7 in oxidation state and decreased to Mn²⁺. As the oxidation state is lower, we talk about the reduction. Therefore, the permanganate is the oxidizing agent because it oxidizes iodide to iodine
