Answer:
The answer is 6.25g.
Explanation:
First create your balanced equation. This will give you the stoich ratios needed to answer the question:
2C8H18 + 25O2 → 16CO2 + 18H2O
Remember, we need to work in terms of NUMBERS, but the question gives us MASS. Therefore the next step is to convert the mass of O2 into moles of O2 by dividing by the molar mass:
7.72 g / 16 g/mol = 0.482 mol
Now we can use the stoich ratio from the equation to determine how many moles of H2O are produced:
x mol H2O / 0.482 mol O2 = 18 H2O / 25 O2
x = 0.347 mol H2O
The question wants the mass of water, so convert moles back into mass by multiplying by the molar mass of water:
0.347 mol x 18 g/mol = 6.25g
Answer: COMBINED FORCES
When forces act in the same direction, they combine to make a bigger force. When they act in opposite directions, they can cancel one another out. If the forces acting on an object balance, the object does not move, but may change shape.
Explanation:
Reduction involves the either the addition of hydrogen and removal of oxygen.
<h3>What is reduction?</h3>
Reduction involves the removal of oxygen.
This implies there is a loss of oxygen in reduction.
This can be represented in the extraction of iron from it ores.
Fe₂O₃ + 3CO → 2Fe + 3CO₂
Reduction is also the addition of hydrogen. This implies it is the gain of hydrogen.
For example
CH₃CHO → CH₃CH₂OH
learn more on reduction here: brainly.com/question/9485345
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Answer:
a) 
b) 
Explanation:
Equation of reaction:
Initial pressure 3 1 0
Pressure change 2P 1P 2P
Total pressure = (3-2P) + (1-P) + (2P)
Total Pressure = 3.75 atm
(3-2P) + (1-P) + (2P) = 3.75
4 - P = 3.75
P = 4 - 3.75
P = 0.25 atm
Let us calculate the pressure of each of the components of the reaction:
Pressure of XO2 = 3 - 2P = 3 - 2(0.25)
Pressure of XO2 =2.5 atm
Pressure of O2 = 1 - P = 1 -0.25
Pressure of O2 = 0.75 atm
Pressure of XO3 = 2P = 2 * 0.25
Pressure of XO3 = 0.5 atm
From the reaction, equilibrium constant can be calculated using the formula:
![K_{p} = \frac{[PXO_{3}] ^{2} }{[PXO_{2}] ^{2}[PO_{2}] }](https://tex.z-dn.net/?f=K_%7Bp%7D%20%3D%20%5Cfrac%7B%5BPXO_%7B3%7D%5D%20%5E%7B2%7D%20%7D%7B%5BPXO_%7B2%7D%5D%20%5E%7B2%7D%5BPO_%7B2%7D%5D%20%7D)
Standard free energy:
b) value of k−1 at 27 °C, i.e. 300K
