The maximum mass of B₄C that can be formed from 2.00 moles of boron (III) oxide is 55.25 grams.
<h3>What is the stoichiometry?</h3>
Stoichiometry of the reaction gives idea about the relative amount of moles of reactants and products present in the given chemical reaction.
Given chemical reaction is:
2B₂O₃ + 7C → B₄C + 6CO
From the stoichiometry of the reaction, it is clear that:
2 moles of B₂O₃ = produces 1 mole of B₄C
Now mass of B₄C will be calculated by using the below equation:
W = (n)(M), where
- n = moles = 1 mole
- M = molar mass = 55.25 g/mole
W = (1)(55.25) = 55.25 g
Hence required mass of B₄C is 55.25 grams.
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Q = m x c x ΔT
2500 = 0.135 x C x 80.5
2500 = 10.8765 x C
C = 230.043 J/Kg.K
hope this helps
1) Zn(CH₃COO)₂(s) + 2KOH(aq) = Zn(OH)₂(s) + 2CH₃COOK(aq)
Ksp{Zn(OH)₂}=1.2*10⁻¹⁷
2) Zn(CH₃COO)₂(s) + 2NaCN(aq) = Zn(CN)₂(s) + 2CH₃COONa(aq)
Ksp{Zn(CN)₂}=2.6*10⁻¹³
Ksp{Zn(OH)₂}<Ksp{Zn(CN)₂}
Zn(OH)₂ precipitates first
Answer:
Explanation:
Combustion reaction is given below,
C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)
Provided that such a combustion has a normal enthalpy,
ΔH°rxn = -1270 kJ/mol
That would be 1 mol reacting to release of ethanol,
⇒ -1270 kJ of heat
Now,
0.383 Ethanol mol responds to release or unlock,
(c) Determine the final temperature of the air in the room after the combustion.
Given that :
specific heat c = 1.005 J/(g. °C)
m = 5.56 ×10⁴ g
Using the relation:
q = mcΔT
- 486.34 = 5.56 ×10⁴ × 1.005 × ΔT
ΔT= (486.34 × 1000 )/5.56×10⁴ × 1.005
ΔT= 836.88 °C
ΔT= T₂ - T₁
T₂ = ΔT + T₁
T₂ = 836.88 °C + 21.7°C
T₂ = 858.58 °C
Therefore, the final temperature of the air in the room after combustion is 858.58 °C
<span>These atoms are known as valence atoms.</span>
