<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.
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Answer:
d. Copper (II) sulfate
Explanation:
Given data:
Mass of Al = 1.25 g
Mass of CuSO₄ = 3.28 g
What is limiting reactant = ?
Solution:
Chemical equation:
2Al + 3CuSO₄ → Al₂ (SO₄)₃ + 3Cu
Number of moles of Al:
Number of moles = mass/molar mass
Number of moles = 1.25 g/ 27 g/mol
Number of moles = 0.05 mol
Number of moles of CuSO₄:
Number of moles = mass/molar mass
Number of moles = 3.28 g/ 159.6 g/mol
Number of moles = 0.02 mol
now we will compare the moles of reactant with product.
Al : Al₂ (SO₄)₃
2 : 1
0.05 : 1/2×0.05=0.025 mol
Al : Cu
2 : 3
0.05 : 3/2×0.05 = 0.075 mol
CuSO₄ : Al₂ (SO₄)₃
3 : 1
0.02 : 1/3×0.02=0.007 mol
CuSO₄ : Cu
3 : 3
0.02 : 0.02
Less number of moles of reactants are produced by CuSO₄ thus it will act as limiting reactant.
Answer:
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2
Explanation:
Hello,
In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.
Moreover, in the other reactions, copper (A.), sodium (C. and D.) remain with the same initial oxidation state, +2 and +1 respectively.
Regards.
