The percent yield of carbon dioxide will be 49.0 %.
<h3>Percent yield</h3>
First, let's look at the equation of the reaction:

The mole ratio of octane to oxygen is 2:25.
Mole of 3.43 g octane = 3.43/114.23 = 0.03 mol
Mole of 19.1 g oxygen = 19.1/32 = 0.60 mol
Thus, octane is limiting.
Mole ratio of octane to carbon dioxide = 2:16.
Equivalent mole of carbon dioxide = 0.03 x 8 = 0.24 mol
Mass of 0.24 mol carbon dioxide = 0.24 x 44.01 = 10.5624 grams
Percent yield of carbon dioxide = 5.18/10.5624 = 49.0 %
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Answer:
n = 3 to n = 5
Explanation:
According to the Bohr's model of the atom, electrons in an atom absorb energy to move from a lower to higher energy level.
We must note that as we progress away from the nucleus, the energy levels of electrons become closer together. The energy difference between successive levels decreases and the wavelength of light associated with such transitions become longer.
Hence,the absorption of light of the longest wavelength corresponds to n = 3 to n = 5
.
When solid carbon reacts with oxygen gas to produce carbon dioxide gas. the deltaH (enthalpy change ) value is negative .DeltaH would be on the product side of the equation.
<h3>What is enthalpy change? </h3>
In a thermodynamic system, energy is measured by enthalpy. Enthalpy is a measure of a system's overall heat content and is equal to the system's internal energy plus the sum of its volume and pressure.
Knowing whether q is endothermic or exothermic allows one to characterise the relationship between q and H. An endothermic reaction is one that absorbs heat and demonstrates that heat from the environment is used in the reaction, hence q>0 (positive). For the aforementioned equation, under constant pressure and temperature, if q is positive, then H will also be positive. In a similar manner, heat is transferred to the environment when it is released during an exothermic reaction. Thus, q=0 (negative). Therefore, if q is negative, H will also be negative.
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Answer:
A) During this procedure ( hypoventilation ) The CO2 in the arterial blood vessels and the lungs increases and this drives the PH level in the system lower, and the equilibrium will shift to the right. this is because the Blood-PH level is controlled by CO2 - bicarbonate buffer system
B) The blood PH may rise to 7.60 during Hyperventilation because the removal of CO2 from the lungs causes the increase in
which is directly proportional to the increase in Blood PH levels
C) Hyper ventilation before a dash would be useful because it will remove excessive Hydrogen ions and and raise the Blood PH levels in preparedness of the production of acids like Lactic acid
Explanation:
A) During this procedure ( hypoventilation ) The CO2 in the arterial blood vessels and the lungs increases and this drives the PH level in the system lower, and the equilibrium will shift to the right. this is because the Blood-PH level is controlled by CO2 - bicarbonate buffer system
⇄ 
B) The blood PH may rise to 7.60 during Hyperventilation because the removal of CO2 from the lungs causes the increase in
which is directly proportional to the increase in Blood PH levels
C) Hyper ventilation before a dash would be useful because it will remove excessive Hydrogen ions and and raise the Blood PH levels in preparedness of the production of acids like Lactic acid