Answer:
0.027 mole of NaOH.
Explanation:
We'll begin by obtaining the number of mole H2SeO4 in 45mL of 0.30M H2SeO4
This is illustrated below:
Molarity of H2SeO4 = 0.3M
Volume of solution = 45mL = 45/1000 = 0.045L
Mole of H2SeO4 =...?
Mole = Molarity x Volume
Mole of H2SeO4 = 0.3 x 0.045
Mole of H2SeO4 = 0.0135 mole
Next, the balanced equation for the reaction. This is given below:
H2SeO4 + 2NaOH –> Na2SeO4 + 2H2O
From the balanced equation above,
1 mole of H2SeO4 required 2 moles of NaOH.
Therefore, 0.0135 mole of H2SeO4 will require = 0.0135 x 2 = 0.027 mole of NaOH.
Therefore, 0.027 mole of NaOH is needed for the reaction.
Answer: Lanthanide or Lanthanides
Explanation: Sorry that I don’t have an explanation. I just answered the question correctly myself. <33
The reaction corresponds to the combustion of propane (C3H8). The balanced reaction is:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
The reaction enthalpy is given as:
ΔHrxn = ∑nΔH°f(products) - ∑ nΔH°f(reactants)
= [3ΔH°f(CO2(g)) + 4ΔH°f(H2O(g)] - [1ΔH°f(C3H8(g)) + 5ΔH°f(O2(g)]
= [3(-393.5) + 4(-241.8)] - [-103.9 + 5(0)] = -2043.8 kJ
The enthalpy for the combustion of propane is -2044 kJ
1) Chemical reaction: 2Al + 3Br₂ → 2AlBr₃.
m(Al) = 3,0 g.
m(Br₂) = 6,0 g.
n(Al) = m(Al) ÷ M(Al).
n(Al) = 3,0 g ÷ 27 g/mol.
n(Al) = 0,11 mol.
n(Br₂) = n(Br₂) ÷ m(Br₂).
n(Br₂) = 6 g ÷ 160 g/mol.
n(Br₂) = 0,0375 mol; limiting reagens.
n(Br₂) : n(AlBr₃) = 3 : 2.
n(AlBr₃) = 0,025 mol.
m(AlBr₃) = 0,025 mol · 266,7 g/mol.
m(AlBr₃) = 6,67 g.
2) m(Br₂) - all bromine reacts, so mass of bromine after reaction is zero grams (m(Br₂) = 0 g).
n(Al) = 0,11 mol - 0,025 mol = 0,085 mol.
m(Al) = 0,085 mol · 27 g/mol.
m(Al) = 2,295 g.
m(AlBr₃) = 6,67 g · 0,72 (yield of reaction).
m(AlBr₃) = 4,8 g.
n - amount of substance.
M - molar mass.
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