We have that the Complete Expanded Structure of (CH3)2CHCH2OCH2CH3 is given in the attachment below
From the Question
(CH3)2CHCH2OCH2CH3
Generally for the condensed formula (CH3)2CHCH2OCH2CH3
We consider that this is a single bond connecting them
We consider
Hydrogen H(1)
Oxygen(8)
Carbon(6)
In conclusion
The Complete Expanded Structure of (CH3)2CHCH2OCH2CH3 is given in the attachment below.
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The answer is:
E per gram = 0.45 V
The explanation:
when MnO2 is the substance who oxidized here so, the oxidizing agent and the anode here is Li.
and when the molar mass of Li is = 7 g/mol
and in our reaction equation we have 1 mole of Li will give 3.15 V of the electrical energy
that means that :
7 g of Li gives → 3.15 V
So 1 g of Li will give→ ???
∴ The E per gram = 3.15 V / 7 g of Li
= 0.45 V
Answer:
5.35m H2O2 x 34.02g/1m H2O2 = 182g H2O2
Explanation:
The best answer among the choices is option D. New substances are being formed will best explain that this process is a chemical change. Although, it cannot be seen directly we knew that an acid and base will undergo neutralization reaction forming a salt and water. The heat given off by the reaction can be a parameter however it does not only happens for reactions. It can also happen when dissolving substances.
Answer:
70mol
Explanation:
The equation of the reaction is given as:
2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
Given parameters:
Number of moles of acetylene = 35.0mol
Number of moles of oxygen in the tank = 84.0mol
Unknown:
Number of moles of CO₂ produced = 35.0mol
Solution:
From the information given about the reaction, we know that the reactant that limits this combustion process is acetylene. Oxygen is given in excess and we don't know the number of moles of this gas that was used up. We know for sure that all the moles of acetylene provided was used to furnish the burning procedure.
To determine the number of moles of CO₂ produced, we use the stoichiometric relationship between the known acetylene and the CO₂ produced from the balanced chemical equation:
From the equation:
2 moles of acetylene produced 4 moles of CO₂
∴ 35.0 mol of acetylene would produced:
= 70mol