Answer:
C) Q < K, reaction will make more products
Explanation:
- 1/8 S8(s) + 3 F2(g) ↔ SF6(g)
∴ Kc = 0.425 = [ SF6 ] / [ F2 ]³
∴ Q = [ SF6 ] / [ F2 ]³
∴ [ SF6 ] = 2 mol/L
∴ [ F2 ] = 2 mol/L
⇒ Q = ( 2 ) / ( 2³)
⇒ Q = 0.25
⇒ Q < K, reaction will make more products
Molar mass :
Li₂S = <span>45.947 g/mol
AlCl</span>₃ = <span>133.34 g/mol
</span><span>3 Li</span>₂<span>S + 2 AlCl</span>₃<span> = 6 LiCl + Al</span>₂S₃
3 * 45.947 g Li₂S ----------> 2 * <span>133.34 g AlCl</span>₃
1.084 g Li₂S ----------------> ?
Mass Li₂S = 1.084 * 2 * 133.34 / 3 * 45.947
Mass Li₂S = 289.08112 / 137.841
Mass Li₂S = 2.0972 g
hope this helps!
Answer:
1.35 g
Explanation:
Data Given:
mass of Potassium Permagnate (KMnO₄) = 3.34 g
Mass of Oxygen: ?
Solution:
First find the percentage composition of Oxygen in Potassium Permagnate (KMnO₄)
So,
Molar Mass of KMnO₄ = 39 + 55 + 4(16)
Molar Mass of KMnO₄ = 158 g/mol
Calculate the mole percent composition of Oxygen in Potassium Permagnate (KMnO₄).
Mass contributed by Oxygen (O) = 4 (16) = 64 g
Since the percentage of compound is 100
So,
Percent of Oxygen (O) = 64 / 158 x 100
Percent of Oxygen (O) = 40.5 %
It means that for ever gram of Potassium Permagnate (KMnO₄) there is 0.405 g of Oxygen (O) is present.
So,
for the 3.34 grams of Potassium Permagnate (KMnO₄) the mass of Oxygen will be
mass of Oxygen (O) = 0.405 x 3.34 g
mass of Oxygen (O) = 1.35 g
Red giants produce "metals", i.e., heavier elements.
The first step is helium conversion into
<span>carbon</span>