Answer:
n = 3.0 moles
V = 60.0 L
T = 400 K
From PV = nRT, you can find P
P = nRT/V = (3.0 mol)(0.0821 L-atm/K-mol)(400 K)/60.0L
P = 1.642 atm = 1.6 atm (to 2 significant figures)
Explanation:
Answer:
The moles of sucrose that are available for this reaction is 0.0292 moles
Explanation:
Combustion is an specifyc reaction where the reactants react with O₂ in order to produce CO₂ and H₂O
This combustion is: C₁₂H₂₂O₁₁ + 12O₂ → 12CO₂ + 11H₂O
We have to conver the mass to moles, to find out the limiting reactant
10 g . 1 mol / 342 g = 0.0292 moles of sucrose
8 g . 1mol / 32g = 0.250 moles of O₂
The moles of sucrose that are available for this reaction is 0.0292 moles
Before we start to work with the equation we must find the limiting reactant. When you find it, you can do all the calculations.
Answer:
B
Explanation:
I had tried my best to get this answer and I had gotten B
sorry if it was not the right answer but I would recommend B
Answer:
The answer to your question is below
Explanation:
I just write the formulas of the reactants and products and balanced the reactions.
a)
3H₂(g) + N₂(g) ⇒ 2NH₃
b)
2K + 2H₂O ⇒ 2KOH + H₂ (g)
c)
2Al(s) + Fe₂O₃ ⇒ Al₂O₃ + 2Fe
d)
Fe₂O₃ + 2Al ⇒ Al₂O₃ + 2Fe
e)
Ba(OH)₂ + 2HBr ⇒ BaBr₂ + 2H₂O
f)
CaCO₃ + Δ ⇒ CaO + CO₂
Answer:
an exothermic reaction that Increase in entropy
