<u>The frequency of </u><u>collisions </u><u>between the two reactants increases as the </u><u>concentration </u><u>of the reactants increases</u>. When collisions happen, they don't always cause a reaction (atoms misaligned or insufficient energy, etc.). Higher concentrations result in more collisions and reaction opportunities.
Increasing a reactant's surface area increases the frequency of collisions and thus the reaction rate. The surface area of several smaller particles is greater than that of a single large particle. The greater the available surface area for particles to collide, the faster the reaction will occur.
<h3>How does concentration affect the rate of collisions between reactants?</h3>
Thus, we can conclude that by increasing the concentration of Mg in the reaction mixture we increase the rate of collisions between the reactants in this reaction.
<h3>What does the half reaction of an oxidation-reduction reaction show?</h3>
Iron gains electrons in the half reaction of an oxidation-reduction reaction. What does iron's electron gain mean? It has been reduced. Predict the product that will precipitate out of the reaction using the solubility rules and the periodic table.
Learn more about collisions of particles:
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Adaptation actually and also following control measures on how to avoid it from happening
Answer:
2C₃H₇BO₃ + 8O₂ → 6CO₂ + 7H₂O + B₂O₃.
Explanation:
- For balancing a chemical equation, we should apply the law of conversation of mass. It states that the no. of atoms in the reactants side is equal to that of the products side.
So, the balanced equation:
<em>2C₃H₇BO₃ + 8O₂ → 6CO₂ + 7H₂O + B₂O₃.</em>
It is clear that 2.0 moles of C₃H₇BO₃ is completely burned in 8 m oles of oxygen and produce 6 moles of CO₂, 7 moles of H₂O and 1 mole of B₂O₃.
According to the balanced equation of the reaction:
2C2H2 + 5O2 → 4CO2 + 2H2O
So we can mention all as liters,
A) as we see that 2 liters of C2H2 react with 5 liters of oxygen to produce 4 liters of CO4 and 2 liters of H2O
So, when we have 75L of CO2
and when we have 2 L of C2H2 reacts and gives 4 L of CO2
2C2H2 → 4CO2
∴ The volume of C2H2 required is:
= 75L / 2
= 37.5 L
B) and, when we have 75 L of CO2
and 4CO2 → 2H2O
∴ the volume of H2O required is:
= 75 L /2
= 37.5 L
C) and from the balanced equation and by the same way:
when 5 liters O2 reacts to give 4 liters of CO2
and we have 75 L of CO2:
5 O2 → 4 CO2
?? ← 75 L
∴ the volume of O2 required is:
= 75 *(5/4)
= 93.75 L
D) about the using of the number of moles the answer is:
no, there is no need to find the number of moles as we called everything in the balanced equation by liters and use it as a liter unit to get the volume, without the need to get the number of moles.
