Answer:
4.5 moles of carbon dioxide
Explanation:
Given parameters:
Mass of CH₄ = 72g
Unknown:
number of moles of CO₂ produced = ?
Solution:
To solve this problem, let us write the reaction equation first;
CH₄ + 2O₂ → CO₂ + 2H₂O
Now we know that O₂ is in excess and CH₄ is the limiting reactant that will determine the extent of the reaction.
Let us find the number of moles of CH₄ ;
Number of moles = 
molar mass of CH₄ = 12 + 4 = 16g/mol
Number of moles = 
= 4.5moles
From the reaction equation;
1 mole of methane produced 1 mole of carbon dioxide
4.5 moles of methane will produce 4.5 moles of carbon dioxide
I think the answer would be : Tetrahedral
Oxygen has 6 electrons and need 2 more electron to complete its octet. In Tetrahedral, the electrons are arranged so it surround the oxygen and forming a 109 degree bond angel/
hope this helps
<span>CaCO3(Calcium Carbonate)</span>
Answer:
The answer to your question is: 16.7 g of KBr
Explanation:
Data
mass KBr = ? g
Volume = 0.400 L
Concentration = 0.350 M
Formula
Molarity = moles / volume
moles = molarity x volume
Process
moles = (0.350)(0.400)
= 0.14
MW KBr = 39 + 80 = 119 g
119 g of KBr -------------------- 1 mol
x -------------------- 0.14 mol
x = (0.14 x 119) / 1
x = 16.7 g of KBr
Responda:
+ 0,9kJ / mol
Explicação:
Dados os calores de combustão do enxofre monoclínico e enxofre rômbico como - 297,2 kJ / mol e - 296,8 kJ / mol, respectivamente para a variação na transformação de 1 mol de enxofre rômbico em enxofre monoclínico conforme mostrado pela equação;
S (mon.) + O2 (g) -> SO2 (g)
Uma vez que são todos 1 mol cada, a mudança na entalpia será expressa como ∆H = ∆H2-∆H1
Dado ∆H2 = -296,8kJ / mol
∆H1 = -297,2kJ / mol
∆H = -296,8 - (- 297,2)
∆H = -296,8 + 297,2
∆H = 297,2-296,8
∆H = + 0,9kJ / mol
Portanto, a mudança na entalpia da equação é + 0,9kJ / mol
