Kc= concentration of product divided by concentration of reactant
NO + NO2 ----> N2O3
Kc =(N2O3) / (No)(NO2)
Kc= ( 1.3 )/{ (3.9)(3.8) }
Kc=0.088 ( answer B)
First blank -Same
Second blank-Neutral
Find the number of moles
C = n / V
C(Concentration) = 0.30 moles / L
V ( Volume) = 2 L
n = ??
n = C * V
n = 0.30 mol / L * 2 L
n = 0.60 mol
Find the molar mass
2Na = 23 * 2 = 46 grams
1S = 32 * 1 = 32 grams
O4 = 16 * 4 = 64 grams
Total = 142 grams / mol
Find the mass
n = given mass / molar mass
n = 0.06 mol
molar Mass = 142 grams / mol
given mass = ???
given mass = molar mass * mols
given mass = 142 * 0.6
given mass = 85.2 grams.
85.2 are in a 2 L solution that has a concentration of 0.6 mol/L
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
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