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ruslelena [56]
3 years ago
15

1. Butane gas (C4H10) combusts with oxygen gas (O2) to form water (H2O) and carbon dioxide (CO2). Write a balanced;’ equation fo

r this reaction and explain the scientific principle that requires the balancing of an equation to make it conform to reality.
Chemistry
1 answer:
Mars2501 [29]3 years ago
5 0

Answer:

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O.

Explanation:

  • To balance a chemical reaction, we apply the law of conversation of mass, states that the no. of atoms in both sides of the reactants and products is the same.

So, the balanced equation of combustion of butane is:

<em>2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O.</em>

  • <em>It is clear that 2.0 moles of butane is burned in 13.0 moles of oxygen to produce 8.0 moles of CO₂ and 10.0 moles of H₂O.</em>
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The correct answer is option C, that is, hypothesis.  

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3 years ago
Does anyone know this
LekaFEV [45]

Answer:

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3 0
3 years ago
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Iodine-131 has a half-life of 8 days. If the amount of iodine-131 in the orginial sample is 8 g, how much iodine-131 will remain
hram777 [196]

1/32

Explanation it stated with half on 8 days then that means you divide 24 and 8 so its 3 and you have to multiply ✖ ½

7 0
3 years ago
Use the tabulated half-cell potentials to calculate the equilibrium constant (K) for the following balanced redox reaction at 25
o-na [289]
The equation Eºcell = 0.0592/n logK must be used to find n and also Eºcell 
2 Al(s) + 3 Mg2+(aq) → 2 Al3+(aq) + 3 Mg(s) Al3+ +3e- --> Al Eº = -1.66 V Mg2+ +2e- -->Mg Eº = -2.37V 
To balance the equation, 6 moles of electrons must be transferred (2 Al and 3 Mg). This will be the value of n in the equation. 
To find Eºcell, you need the reduction potentials which should be given in a table, and given above. Eºcell = -1.66 - (-2.37) = 0.71 V log K = Eºcell x n/0.0592 = 0.71 x 6/0.0592 log K = 71.95 K = 10^71.95 K = 1.1x10^72
6 0
3 years ago
It has been suggested that hydrogen gas obtained by the decomposition of water might be a substitute for natural gas (principall
Aleks [24]

Answer:

Hydrogen: -141 kJ/g

Methane: -55kJ/g

The energy released per gram of hydrogen in its combustion is higher than the energy released per gram of methane in its combustion.

Explanation:

According to the law of conservation of the energy, the sum of the heat released by the combustion and the heat absorbed by the bomb calorimeter is zero.

Qc + Qb = 0

Qc = -Qb  [1]

We can calculate the heat absorbed by the bomb calorimeter using the following expression.

Q = C . ΔT

where,

C is the heat capacity

ΔT is the change in the temperature

<h3>Hydrogen</h3>

Qc = -Qb = -C . ΔT = -(11.3 kJ/°C) . (14.3°C) = -162 kJ

The heat released per gram of hydrogen is:

\frac{-162kJ}{1.15g} =-141 kJ/g

<h3>Methane</h3>

Qc = -Qb = -C . ΔT = -(11.3 kJ/°C) . (7.3°C) = -82 kJ

The heat released per gram of methane is:

\frac{-82kJ}{1.50g} =-55kJ/g

3 0
3 years ago
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