Answer:
Cdiasuu alprguta bonrradse??
M = n / V
Where, M is molarity (M or mol/L), n is number of moles of the solute (mol) and V is volume of the solution (L).
Here the solute is KNO₃.
The given molarity is 1.3 M
This means 1L of solution has 1.3 moles of KNO₃.
Hence moles in 600 mL = 1.3 M x 0.6 L = 0.78 mol
Therefore to make 1.3 M KNO₃ solution, needed moles of KNO₃ is 0.78 mol
Answer:
CH4+2O2→CO2+2H2O
Moles of oxygen=3232=1
Moles of carbon dioxide=4422=0.5
Moles of water=1818=1
1 mole of methane reacts with 2 moles of oxygen to give 1 mole of CO2 and 2 moles of water so 0.5 mole of methane will be required which is equal to 0.5×16=8g.
Explanation:
8g
Answer:
A) ΔG° = -3,80x10⁵ kJ
B) E° = 2,85V
Explanation:
A) It is possible to answer this problem using the standard ΔG's of formation. For the reaction:
Mg(s) + Fe²⁺(aq) → Mg²⁺(aq) + Fe(s)
The ΔG° of reaction is:
ΔG° = ΔGFe(s) + ΔGMg²⁺(aq) - (ΔGFe²⁺(aq) + ΔGMg(s) <em>(1)</em>
Where:
ΔGFe(s): 0kJ
ΔGMg²⁺(aq): -458,8 kJ
ΔGFe²⁺(aq): -78,9 kJ
ΔGMg(s): 0kJ
Replacing in (1):
ΔG° = 0kJ -458,8kJ - (-78,9kJ + okJ)
<em>ΔG° = -3,80x10² kJ ≡ -3,80x10⁵ kJ</em>
B) For the reaction:
X(s) + 2Y⁺(aq) → X²⁺(aq) + 2Y(s)
ΔG° = ΔH° - (T×ΔS°)
ΔG° = -629000J - (298,15K×-263J/K)
ΔG° = -550587J
As ΔG° = - n×F×E⁰
Where n are electrons involved in the reaction (<em>2mol</em>), F is faraday constant (<em>96485 J/Vmol</em>) And E° is the standard cell potential
Replacing:
-550587J = - 2mol×96485J/Vmol×E⁰
<em>E° = 2,85V</em>
I hope it helps!
